The ABC Formula

You already know the importance of your A-B-C’s, but you’re probably just now learning the value of the ABC formula. Unfortunately, there’s no catchy song for this ABC (but if you write one, let us know!).

It’s really not as difficult as you think, especially once you break it down into parts.

Let’s dive in!

What is the ABC formula?

The ABC formula is — you guessed it — a formula, specifically used for solving quadratic equations. The name “ABC” comes from the coefficients of the quadratic equation, written in standard form:

$$ax^2+bx+c=0 $$

Once the coefficients are identified, they can be substituted into this formula to find the final solution:


Why is the ABC formula so useful?

The ABC formula provides a more streamlined and learner-friendly way to solve quadratic equations.

Here’s what makes it a great method:

Your goal while solving an equation is to isolate the variable on one side. That’s easy when the variable (AKA the unknown) is not raised to any power — but that’s not the case in quadratic equations, where the variable is raised to the second power.

Other methods of solving quadratic equations (factoring, completing the square, etc.) require much more manipulation and can therefore get very messy, very quickly.

The ABC formula, however, is much cleaner and more efficient. All you need to do is identify the coefficients, insert them into the formula, and evaluate the expression!

See why we’re such big fans?

How to use the ABC formula

Now that we know what the ABC formula is and why it’s useful, it’s time to see it in action! Let’s walk through a problem together.


We’ll solve this quadratic equation using the ABC formula:


Write the equation in standard form, $$ax^2+bx+c=0$$. This helps us easily identify the coefficients $$a$$, $$b$$, and $$c$$:


Identify the coefficients $$a$$, $$b$$, and $$c$$ of the quadratic equation. The coefficient $$a$$ is the one multiplying $$x^2$$, the coefficient $$b$$ is multiplying $$x$$, and the coefficient $$c$$ is the standalone constant.


Now, we’ll substitute $$a=2$$, $$b=-7$$, and $$c=3$$ into the quadratic formula $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-(-7)\pm \sqrt{(-7)^2-4\cdot2\cdot3}}{2\cdot2}$$

All that’s left now is to simplify the expression!

To start, let’s remove the $$-$$ sign from in front of the expression in parentheses. Change the sign of each term of the expression and remove the parentheses.

$$x=\frac{7\pm \sqrt{(-7)^2-4\cdot2\cdot3}}{2\cdot2}$$

Now let’s evaluate the power. Remember: a negative base raised to an even power equals a positive.

$$x=\frac{7\pm \sqrt{49-4\cdot2\cdot3}}{2\cdot2}$$

Calculate the product of the numbers under the square root and in the denominator.

$$x=\frac{7\pm \sqrt{49-24}}{4}$$

Subtract the numbers under the square root.

$$x=\frac{7\pm \sqrt{25}}{4}$$

Time to take the square root!

$$x=\frac{7\pm 5}{4}$$

Write the solutions, one with a $$+$$ sign and one with a $$-$$ sign.

$$x=\frac{7+ 5}{4}$$

$$x=\frac{7- 5}{4}$$

Now we’ll just complete basic operations to calculate our final solutions.



There they are — our final solutions! As you can see, the equation has two solutions. We can index them if we want, like this:


That wasn’t so bad, right? Now that we’ve walked through a detailed example, let’s review the overall process so you can learn how to use it with any problem:

Study summary

  1. Rewrite the quadratic equation in standard form (if it’s not already).
  2. Identify the coefficients.
  3. Insert the coefficients into the ABC formula.
  4. Simplify the expression to get the solutions of your quadratic equation.

Do it yourself!

You might not think practicing math is the most glamorous way to spend your time, but the repetition really helps cement the method in your mind, especially with a memorizable process like the ABC formula. So, when you’re ready, we’ve got some practice problems for you!

Solve the following quadratic equations using the ABC formula:

  1. $$x^2+x-30=0$$
  2. $$-2t^{2} + 3t + 2 = 0$$
  3. $$x^2=25$$
  4. $$3x^2+x=0$$


  1. $$x_1=-6, x_2=5$$
  2. $$t_1=-\frac{1}{2}, t_2=2$$
  3. $$x_1=-5, x_2=5$$
  4. $$x_1=-\frac13, x_2=0$$

If you’re still struggling through the solving process, that’s totally okay! Stumbling a few times is actually good for learning. If you get too stuck or lost, scan the problem using your Photomath app, and we’ll walk you through to the other side!

Here’s a sneak peek of what you’ll see:


Extra credit

Context is so important in getting a 360-degree learning experience. If you’re curious for more, learn how mathematicians originally derived the quadratic formula (spoiler alert: it’s super cool!).

Got algebra homework?

Head over to the Photomath app for instant, step-by-step solutions to all of your algebra problems.