Solving Problems using the Quadratic Formula

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: How to Add and Subtract Rational Expressions

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
Your next lesson will play in 10 seconds
  • 0:05 The Quadratic Formula
  • 0:59 Two Solutions
  • 4:30 One Repeated Solution
  • 6:15 No Real Solutions
  • 7:28 Lesson Summary
Save Save Save

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Log in or Sign up

Speed Speed
Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer

Amy has a master's degree in secondary education and has taught math at a public charter high school.

The quadratic formula is not as complicated as it may look. You can learn to tackle this formula and succeed in using it to help you solve your quadratic problems. Watch this video to see how.

Quadratic Formula

I used to hate working with the quadratic formula until I realized that it actually helped me with my problems. And sometimes, it even made my problem shorter as it gave me what I needed to know early on in the solving process. Right now, the quadratic formula might look like a beast, but think of it as a prince in disguise. It's here to help you. The quadratic formula allows you to solve quadratic equations in the form of ax^2 + bx + c = 0 and is defined as x = (-b +/- sqrt(b^2 - 4ac)) / 2a.

The a, b, and c letters come from the quadratic equation that you are solving. Let's see how this works.

Two Solutions

The first problem we're going to solve is x^2 + 5x + 6 = 0. How do we figure out our letters? Well, we compare this to the general form of a quadratic: ax^2 + bx + c. A good way to do this is to write the general form above your quadratic so you can see how everything lines up. What we are looking for is the location of our x^2, our x, and our number by itself. We can then figure out what our letters stand for.

In our general form, the number attached to the x^2 is the letter a. So, looking at my quadratic, I see that there is nothing in front of the x^2, which means there is a 1 there, so that means my a = 1. Good.

Next, I look for what is front of the x. I see a 5, so that means my b = 5. The number by itself is a 6, so my c = 6. Now I have all my letters. If, though, you have missing terms, then that means the letter that matches that missing term is equal to zero. At this point, the quadratic formula turns into a prince to help you. To solve, all you need to do is plug in all your letters where they belong and evaluate the formula.

Let's do this.

We plug in our numbers to get x = (-5 +/- sqrt(5^2 - 4*1*6)) / 2*1. We begin evaluating by first calculating what is inside the square root.

x = (-5 +/- sqrt(25 - 24)) / 2*1

x = (-5 +/- sqrt (1)) / 2*1

Look at that. We have a 1 inside the square root. It is a positive number inside the square so we know we can calculate that down. If we know the square root off the top of our head, we can go ahead and write that number down. Otherwise, you can use a calculator to find the exact square root. In our case, we know that the square root of 1 is 1, so we'll write that down.

x = (-5 +/- 1) / 2*1

Our formula includes a +/-, so we have to split our problem into two problems, one for the plus and one for the minus. So:

x = (-5 + 1) / 2*1 and x = (-5 - 1) / 2*1

Now, we can finish solving these two little problems.

x = -4 / 2 and x = -6 / 2


x = -2 and x = -3

And yes, we have two solutions. Most times, you will end up with two solutions like this. If they aren't whole numbers, then they'll be decimal numbers. The way you can tell that you have two solutions like this is if you get a positive number inside your square root.

There are two more cases that will give you different numbers of solutions. Let's see what happens when the square root ends up as a 0.

One Repeated Solution

This time, our problem is x^2 + 4x + 4 = 0. Comparing this to our general form of ax^2 + bx + c, we see that our a = 1, our b = 4, and our c = 4. Plugging these values into our quadratic formula gives us x = (-4 +/- sqrt(4^2 - 4*1*4)) / 2*1. Evaluating the square root, we find that it equals zero.

x = (-4 +/- sqrt(16 - 16)) / 2*1

x = (-4 +/- sqrt(0)) / 2*1

To unlock this lesson you must be a Member.
Create your account

Register to view this lesson

Are you a student or a teacher?

Unlock Your Education

See for yourself why 30 million people use

Become a member and start learning now.
Become a Member  Back
What teachers are saying about
Try it risk-free for 30 days

Earning College Credit

Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it risk-free for 30 days!
Create an account