The quadratic formula gives us a reliable way to solve any quadratic equation, including ones that are difficult to factor. But because the formula has several layers, small mistakes can easily slip in and change the final answer.
You may set up the problem correctly and still lose track of a negative sign or square the wrong value. We know how frustrating it can be to lose points on a problem you knew how to solve.
So, our tutors will walk you through the most common mistakes to watch for and show you how to avoid them as you solve quadratic equations.
To solve a quadratic equation, we’re looking for the value, or values of x that make the equation true. Sometimes we can find those values by factoring. When factoring does not work neatly, the quadratic formula gives us another reliable method.
The formula is derived from a quadratic equation in standard form, but we’ll keep this review brief and focus on the parts you need.
First, let’s look at the standard form:
ax² + bx + c = 0.
Here:
a is the coefficient of x². The value of a cannot be 0, because then we would no longer have a quadratic equation.
b is the coefficient of x.
These same values, a, b, and c, appear in the quadratic formula:
x = \(\Large\frac{−b ± \sqrt{(b^2 − 4ac)}}{2a}\).
We call the expression under the square root, b² − 4ac, the discriminant and write it as Δ:
Δ = b² − 4ac.
That means we can also write the quadratic formula as:
x = \(\Large\frac{−b ± \sqrt{Δ}}{2a}\).
The discriminant tells us what kind of solutions to expect:
Δ > 0 means we get 2 distinct real solutions.
Δ = 0 means we get 1 real solution.
Δ < 0 means we get 2 complex solutions and no real solutions.
Now, we’ll put the formula into practice and solve:
x² + 2x − 2 = 0
Step 1: Try factoring. We would need two integers that multiply to −2 and add to 2, but no such pair exists. Since this equation does not factor neatly using integers, we use the quadratic formula.
Step 2: Find the discriminant. In our equation, a = 1, b = 2, c = −2. We substitute these values into the discriminant formula: Δ = b² − 4ac = 2² − 4 × 1 ×(−2) = 12.
Step 3: Identify the number of solutions. The discriminant equals 12, Δ > 0, which tells us that the equation will have 2 distinct real solutions.
Step 4: Substitute the values into the quadratic formula. In our equation, a = 1, b = 2, c = −2. We substitute them in the formula: x = \(\Large\frac{−2 ± \sqrt{12}}{ 2×1}\) = −1 ± √3.
Step 5: Simplify the result if needed. First, we simplify the square root. Since 12 = 4 × 3: √12 = √4 × √3 = 2√3. So, we can write our solution as: x = \(\Large\frac{(−2 ± 2\sqrt{3})}{2}\). Now every term in the numerator is divisible by 2, and we can divide each term by 2, which leaves us: x = −1 ± √3.
So, our two solutions are:
x1 = −1 + √3.
x2 = −1 − √3.
The formula packs several operations into one line, from signs and squares to a square root and a division. So, next, we’ll break down the places where mistakes tend to happen so you know what to watch for.
At Mathnasium, we’ve noticed the same handful of mistakes tend to show up when students first start using the quadratic formula. To make them easier to spot, our tutors explained the most common ones below.
The quadratic formula is built specifically for equations written as ax² + bx + c = 0. So before we pick out a, b, and c, we need to make sure every term is on one side of the equal sign and 0 is on the other. Otherwise, we can easily assign the wrong values.
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Example: x² + 3x = 5. |
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Common mistake |
When students read the equation as it’s written, they might grab a = 1, b = 3, and c = 5, and substitute those into the formula. If we use c = 5, we end up solving a different equation, x² + 3x + 5 = 0, instead of the one we started with. |
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What to do instead |
Rewrite the equation in a standard form first. Move 5 to the left side: x² + 3x − 5 = 0. Now the equation matches standard form, so we can correctly identify: a = 1, b = 3, and c = −5. |
Before reaching for the quadratic formula at all, we recommend running through a quick method check:
Is the equation in standard form, ax² + bx + c = 0? If not, rearrange it first.
Does it factor cleanly, with an integer pair that multiplies to ac and adds to b? If so, factoring is often faster.
If it doesn’t factor neatly, reach for the quadratic formula.
If a = 1 and the middle term is even, you can complete the square, which can also be a quick route.
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Even after we rewrite the equation in standard form, we still need to match each number to the right part of the formula. Negative signs can be especially easy to miss, so we should identify a, b, and c carefully before we substitute their values into the formula.
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Example: 2x² − 7 = 4x. |
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Common mistake |
Before an equation is in standard form, you may read a = 2, b = −7, and c = 4, even though those values do not line up with the x² term, the x term, and the constant in ax² + bx + c = 0. |
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What to do instead |
As you rewrite the equation in standard form, make sure each coefficient stays with the correct term and follows the usual order: x² term, x term, then constant. Our example in standard form looks like this: 2x² − 4x − 7 = 0, which lets us see clearly that a = 2, b = −4, and c = −7.
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To keep coefficients straight, use a coefficient audit:
Write the equation in standard form.
List a = _, b = _, c = _with signs included.
Learners may see (−b) and accidentally use b with its original sign from the equation, instead of changing it to the opposite sign as (−b) tells us to do.
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Example: x² − 5x + 6 = 0. |
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Common mistake |
You might read b as 5 because you focus on the number and overlook the minus sign in front of the x-term. That leads to −b = −5. |
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What to do instead |
Read the coefficient together with its sign: b = −5. Then take the opposite: −b = −(−5) =5. |
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Inside the discriminant, b² − 4ac, two different negative signs can show up at once. One may come from the value of c, while the other comes from the subtraction already built into the formula. We need to keep them separate.
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Example: x² + 4x − 3 = 0. |
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Common mistake |
Students can calculate 4ac = −12 correctly, and b² = 4² = 16, then lose track of the subtraction in the discriminant. That can turn 16 − (−12) into 16 − 12 = 4, even though subtracting a negative should give 16 + 12 = 28. |
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What to do instead |
Work through the discriminant in pieces: b² = 16, 4ac = 4 × 1 ×(−3) = −12, so b² − 4ac = 16 − (−12) = 16 + 12 = 28. |
Here is how you can to catch the mistake early:
After solving, compare what you found to what the discriminant predicted. If Δ turned out negative but your final answer shows two real numbers, that’s a sign to go back and recheck the subtraction.
When b is negative, we need to carry that sign with the number before we square it. In this part of the formula, b² should always be non-negative.
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Example: x² − 3x + 9 = 0. |
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Common mistake |
A common slip is to identify b = −3 correctly, then write −3² = −9. Without parentheses, the exponent applies to 3 first, so the negative sign stays outside the square. |
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What to do instead |
Put the negative value in parentheses before you square it: (−3)² = (−3) × (−3) = 9. That keeps the sign with b and gives the correct value for b². |
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The denominator 2a applies to the entire numerator, so both (−b) and the square-root term need to stay together above the fraction bar.
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Example: x² + 3x − 2 = 0. |
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Common mistake |
After finding the discriminant Δ = 17, the answer may end up written as x = −3 ± \(\Large\frac{\sqrt{17}}{2×1}\). That leaves −3 outside the division and applies the denominator only to √17. |
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What to do instead |
Keep both parts of the numerator grouped together: x = \(\Large\frac{(−3 ± \sqrt{17})}{2×1}\). The fraction bar divides the entire expression −3 ± √17 by 2. |
Quadratic equations can have more than one solution, so we need to be careful with shortcuts that can accidentally remove one before we finish.
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Example: x² = 5x. |
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Common mistake |
It can be tempting to divide both sides by x and get x = 5. But that step assumes x ≠ 0, so we lose x = 0 as a possible solution. |
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What to do instead |
Move everything to one side first: x² − 5x = 0. Then factor: x(x − 5) = 0. From there, we get both solutions: x = 0 or x = 5. |
Avoid dividing by a variable when solving an equation. If you do divide by a variable at some point, go back and check whether that variable could have equaled zero.
Even a correctly solved equation can hand back a root that doesn’t make sense once we return to the original problem. We can catch this kind of slip by checking whether the final answer makes sense.
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Example: A rectangle problem gives two possible side lengths of (−3) and 5 |
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Common mistake |
Both values may come out of the calculations, and at first glance, (−3) and 5 can both look like possible answers. But before we keep them, we need to check what those numbers represent in the problem. |
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What to do instead |
A side length cannot be negative, so (−3) does not fit the situation. That leaves 5 as the meaningful solution. |
In real-world setups, like lengths, time, or number of physical objects, a negative or zero value is impossible, even when it’s a perfectly valid algebraic solution.
Need a quick refresh before a quiz or homework assignment? Use this table to jog your memory on where each mistake tends to happen and how to catch it fast.
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Mix-up |
How to fix |
10-second self-check |
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Setting up the equation incorrectly. |
Rewrite the equation as ax² + bx + c = 0, then list a, b, and c with their signs before you substitute. |
Is one side equal to 0, and do a, b, c match the correct terms? |
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Forgetting to change the sign in (-b). |
Read the coefficient together with its sign and remember that (−b) means the opposite of b. |
Did you change b to its opposite sign? |
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Making sign errors in the discriminant. |
Calculate b² and 4ac separately. Put a negative b in parentheses before squaring, then subtract 4ac carefully. |
Is b² non-negative, and did you keep the minus in b²−4ac? |
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Dividing only part of the numerator. |
Draw a fraction bar under the entire numerator. |
Is 2a under both −b and √Δ? |
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Losing one of the solutions. |
Keep both ± cases when you use the quadratic formula, and avoid dividing by a variable when factoring because that can remove x = 0 as a possible solution. |
Did you keep every possible solution from the formula or factoring? |
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Keeping an answer that does not fit the problem. |
Go back to the original context and check what each solution represents. |
Does each answer make sense for the situation? |
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Each solution below is only partly worked out, and each one contains one mistake. See if you can find it. You can check the answers at the bottom of the page.
Solve x² − 6x + 5 = 0:
a = 1, b = −6, c = 5.
Discriminant: Δ = (−6)² + 4 × 1 × 5 = 36 + 20 = 56.
x = \(\Large\frac{(6 ± \sqrt{(56)}}{2}\).
Solve 2x² + 4x − 3 = 0:
a = 2, b = 4, c = −3.
Discriminant: Δ = 4² − 4 × 2 × (−3) = 16 + 24 = 40.
x = −4 ± \(\Large\frac{\sqrt{40}}{2×2}\) = −4 ± \(\Large\frac{\sqrt{40}}{4}\).
By giving students time to think through mistakes and guiding the reasoning instead of giving answers, Mathnasium tutors help build solid problem-solving skills and independence in math.
Mathnasium is a math-only learning center dedicated to helping K–12 students of all skill levels excel in math.
We make sure students truly understand why methods and formulas work instead of simply memorizing them, so they can apply their skills to unfamiliar problems and catch their own mistakes.
To support that understanding, we use the Mathnasium Method™, our proprietary teaching approach, to meet students where they are and guide them forward step by step.
Each student begins with a diagnostic assessment that helps us understand their current skill level, knowledge gaps, goals, and how they think and feel about math.
Using these insights, we build a personalized learning plan focused on the skills the student needs most, whether that means reinforcing earlier algebra foundations, using the quadratic formula accurately, checking solutions, or preparing for more advanced algebra.
Our specially trained tutors follow that plan closely and provide live, face-to-face instruction in a caring and fun group environment. They use mental, verbal, visual, tactile, and written techniques to help students identify coefficients correctly, substitute them into the formula, keep track of negative signs, simplify the discriminant, and work through each step with care.
Students also get room to think through problems and review their own mistakes before tutors step in. Our tutors guide them through the reasoning rather than simply giving the correct answers. This helps students build critical thinking, problem-solving skills, and greater independence in algebra and beyond.
Fun is part of the approach, too. We use game-based activities, rewards, and consistent encouragement to keep students engaged as they revisit earlier skills and work through more advanced algebra topics.
Families see the difference:
94% of parents report an improvement in their child's math skills and understanding
93% of parents report their child's improved attitude toward math after attending Mathnasium
90% of students saw an improvement in their school grades
With over 1,100 learning centers across North America, there is likely a Mathnasium close to you.
For families in and near Naples, Florida, Mathnasium of Naples brings that same approach close to home, with specially trained tutors who help students make sense of the quadratic formula, avoid common errors, and build the algebra skills that support later coursework.
If your teen is finding the quadratic formula or any other math topic difficult, or keeps losing points to avoidable mistakes, a free diagnostic assessment is a great place to start. Using what we learn, we create a personalized learning plan focused on the skills they need next.
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Did you catch the mix-ups? Here’s what went wrong in each practice problem and how to fix it.
The mistake here is a sign error in the discriminant. The discriminant should be b² − 4ac instead of b² + 4ac. So the correct discriminant is:
(−6)² − 4 × 1 × 5 = 36 − 20 = 16.
That gives us x = \(\Large\frac{(6 ± \sqrt{16})}{2} = \Large\frac{(6 ± 4)}{2}\), so x = 5 or x = 1.
Here, the fraction bar does not extend far enough. Both −4 and ±√40 belong in the numerator, so the entire expression needs to be divided by 4. The correct simplification is:
x = \(\Large\frac{(−4 ± \sqrt{40})}{4}\).
Next, we simplify √40:
√40 = √(4 × 10) = 2√10.
So, our equation can be written as:
x = \(\Large\frac{(−4 ± 2\sqrt{10})}{4}\).
Now every term in the numerator and the denominator is divisible by 2:
x = \(\Large\frac{(−2 ± \sqrt{10})}{2}\).
Mathnasium of Naples is a math-only learning center for K-12 students in Naples, FL. Trusted by over a million parents, Mathnasium uses personalized learning plans and the proprietary Mathnasium Method™ to help students catch up, keep up, and get ahead on their math journey.
Our specially trained tutors deliver face-to-face instruction in a supportive and fun small-group environment, working with students to develop a deep understanding of math, build confidence, and improve academic performance.
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