Quadratic Formula vs. Factoring: What to Choose When Solving Quadratic Equations?

When you solve a quadratic equation, factoring and the quadratic formula can both lead you to the solutions. Each method has its own strengths, and the best choice depends on the equation in front of you.

The quadratic formula works for any quadratic equation, but it can take longer and leave more room for arithmetic errors. Some equations, though, factor in just a few steps.

So how do you choose? Today, we’ll look at the clues in a quadratic equation and learn how to pick the method that gets you to the solution most efficiently.

Review: How to Use Factoring and the Quadratic Formula

When we solve a quadratic equation, we’re looking for the value, or values, of x that make the equation true. Factoring and the quadratic formula can both get us there, but they work in different ways. 

Let’s quickly review both.

1. Factoring

Factoring allows us to rewrite a quadratic expression as the product of two simpler expressions. To factor a quadratic equation, we follow these steps: 

• Step 1: Write the equation in standard form, ax² + bx + c = 0.

• Step 2: Find two numbers that fit the coefficients in the quadratic and let us break the expression into two factors.

• Step 3: Use those numbers to write the quadratic as two factors.

• Step 4: Set each factor equal to 0 and solve for x.

Here’s how it works.

Take this equation:

x² + 5x + 6 = 0.

We need two numbers that multiply to 6 and add to 5. So we start with the factor pairs of 6 and check which pair also gives us a sum of 5:

• 1 and 6 multiply to 6, but they add to 7.

• 2 and 3 multiply to 6 and add to 5.

So 2 and 3 are the pair that works. We can rewrite the quadratic as: 

(x + 2)(x + 3) = 0.

Now we use the zero-product property, which means that when the two factors multiply to 0, at least one of them must equal 0:

x + 2 = 0 or x + 3 = 0.

Next, we solve each equation:

1. x + 2 = 0 → x + 2 - 2 = 0-2 → x = −2.

2. x + 3 = 0 → x + 3 - 3 = 0 - 3 → x = −3.

Our two solutions are -2 and -3.

But what do we do when we cannot find a pair of integers that works? That’s where the quadratic formula comes in handy. 

2. Quadratic Formula

When factoring does not give us a clean pair of factors, we can turn to the quadratic formula instead:

x = \(\Large\frac{−b ± \sqrt{(b^2 − 4ac)}}{2a}\)

Unlike factoring, the quadratic formula does not ask us to search for a pair of numbers that fits the equation. We only need the values of a, b, and c from ax² + bx + c = 0. Before we compare the two methods, we’ll refresh what each part of the formula represents. 

• a, b, and c are the coefficients from the equation in standard form: a is the coefficient of x², b is the coefficient of x, c is the constant term.

• ± tells us to work out two cases, one with addition and one with subtraction.

• b² −4ac is the discriminant, we write it as Δ. It tells us what kind of solutions to expect: Δ > 0 gives 2 real solutions, Δ = 0 gives 1 repeated real solution, and Δ < 0 gives no real solutions.

Because Δ stands for b² −4ac, we can also write the quadratic formula as:

x = \(\Large\frac{−b ± \sqrt{Δ}}{2a}\).

Now, let’s put the formula into practice and work through the steps with this example:

x² + 4x + 1 = 0.

• Step 1: Write the equation in standard form, ax² + bx + c = 0. Our example is already in standard form.

• Step 2: Identify a, b, and c. Here, a = 1, b = 4, c = 1.

• Step 3: Calculate the discriminant: b² − 4ac = 4² − 4 × 1 × 1 = 16 − 4 = 12.  As 12 is positive, we know we’re getting two real solutions. 

• Step 4: Substitute the values into the formula: x = \(\Large\frac{−4 ± \sqrt{12}}{2×1} = \Large\frac{−4 ± \sqrt{12}}{2}\).

• Step 5: Simplify if needed. We rewrite the √12 as √4 × √3, which gives us 2√3. Now, our answer looks like this: x = \(\Large\frac{−4 ± 2\sqrt{3}}{2}\). Each term in the numerator can be divided by 2: x = -2 ± √3.

Our two solutions are x = −2 + √3 and x = −2 − √3.

Factoring vs. Quadratic Formula: What to Choose

In our experience, students often find it tricky to choose the right method when they first start working with quadratic equations. So, our tutors built a quick table to help you assess an equation and decide which method to try first.

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Use Discriminant as a Decision Aid

We can also use the discriminant as a clue about which method may work best. Here’s what to look for: 

• Δ is a perfect square, such as 0, 1, 4, 9, or 16: The quadratic may factor neatly, so factoring is a good place to start.

• Δ is positive but not a perfect square: The solutions will be irrational, so we can go straight to the quadratic formula.

• Δ equals 0: We get one repeated real solution. In this case, the equation factors into two identical factors, so we may try factoring first.

• Δ is less than 0: There are no real solutions, so factoring over the real numbers will not work. Use the quadratic formula instead.

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Your Turn! How Would You Solve These Quadratic Equations?

Try using the decision table and the discriminant to decide which method you’d try first and why. You may check the answers at the bottom of the page.

1. x² − 7x + 10 = 0.

2. 3x² + 2x − 1 = 0.

3. 2x² − 5x − 4 = 0.

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At Mathnasium, we use a diagnostic assessment to spot each student’s learning needs, and from there guide them toward math mastery step by step. 

How Mathnasium Helps Students With Quadratic Equations (and Beyond)

Mathnasium is a math-only learning center dedicated to helping K–12 students of all skill levels excel in math.

Whether students are learning to factor a quadratic equation, use the quadratic formula, or choose between the two, our specially trained algebra tutors help them see the why behind the how and recognize when one method may be more efficient than another. 

To do that, 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 factoring, using the quadratic formula accurately, comparing solution methods, or preparing for more advanced algebra.

Our 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 notice equation structure, compare possible methods, work through each solution carefully, and connect the procedure to the algebra behind it.

Students also get room to think through a problem before tutors step in. Our tutors guide them to explain why they chose factoring or the quadratic formula, check whether the method fits the equation, and verify whether their solutions make sense. This helps students build critical thinking, problem-solving skills, and greater independence in algebra.

Fun is part of the approach, too. We use game-based activities, rewards, and consistent encouragement to keep students engaged as they revisit earlier algebra skills and take on more advanced math.

The results speak for themselves:

• 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 Clear Lake, Texas, Mathnasium of Clear Lake brings that same approach close to home, with specially trained tutors who help students make sense of quadratic equations, choose effective solution methods, and build the algebra skills that later coursework depends on.

If your teen needs support with quadratic equations or any other math topic, 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.

📅 Schedule a Free Assessment at Mathnasium of Clear Lake

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Check Your Answers Here!

Ready to see how you did? Here’s the reasoning behind each choice, followed by the solutions.

Task 1: 

x² − 7x + 10 = 0 → x = 2 or x = 5. 

Small integer coefficients, and the discriminant, 49 − 40 = 9, is a perfect square. 

Factoring works cleanly: (x−2)(x −5) = 0. 

Then we set each factor equal to zero:  x − 2 = 0 or x − 5 = 0, which gives us x = 2 or x = 5. 

Task 2:

3x² + 2x − 1 = 0 → x = \(\Large\frac{1}{3}\) or x = −1. 

Even with a leading coefficient other than 1, the discriminant, 4 + 12 = 16, is a perfect square, which means the equation factors nicely: (3x −1)(x +1) = 0. 

Then we set each factor equal to zero: 3x − 1 = 0 or x + 1 = 0, which leaves us with x = \(\Large\frac{1}{3}\) or x = −1. 

Task 3:

2x² − 5x − 4 = 0 → x = \(\Large\frac{(5 ± \sqrt{57})}{4}\). 

Here, ac = 2 × (−4) = −8, and there is no integer pair that multiplies to −8 and adds to −5, so factoring does not work neatly. 

The Δ is 25 + 32 = 57, which is positive but not a perfect square. 

That tells us to expect irrational solutions, so the quadratic formula is the more efficient choice: x = \(\Large\frac{(5 ± \sqrt{57})}{4}\). 

Since √57 cannot be simplified, these are our two solutions.