If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Quadratic formula review

The quadratic formula allows us to solve any quadratic equation that's in the form ax^2 + bx + c = 0. This article reviews how to apply the formula.

What is the quadratic formula?

The quadratic formula says that
x, equals, start fraction, minus, start color #e07d10, b, end color #e07d10, plus minus, square root of, start color #e07d10, b, end color #e07d10, squared, minus, 4, start color #7854ab, a, end color #7854ab, start color #e84d39, c, end color #e84d39, end square root, divided by, 2, start color #7854ab, a, end color #7854ab, end fraction
for any quadratic equation like:
start color #7854ab, a, end color #7854ab, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #e84d39, c, end color #e84d39, equals, 0

Example

We're given an equation and asked to solve for q:
0, equals, minus, 7, q, squared, plus, 2, q, plus, 9
This equation is already in the form a, x, squared, plus, b, x, plus, c, equals, 0, so we can apply the quadratic formula where a, equals, minus, 7, comma, b, equals, 2, comma, c, equals, 9:
q=b±b24ac2aq=2±224(7)(9)2(7)q=2±4+25214q=2±25614q=2±1614q=2+1614  ,  q=21614q=1            ,  q=97\begin{aligned} q &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ q &= \dfrac{-2 \pm \sqrt{2^{2} - 4 (-7) (9)}}{2(-7)} \\\\ q &= \dfrac{-2 \pm \sqrt{4 +252}}{-14} \\\\ q &= \dfrac{-2 \pm \sqrt{256}}{-14} \\\\ q &= \dfrac{-2 \pm 16}{-14} \\\\ q &= \dfrac{-2 + 16}{-14} ~~,~~ q = \dfrac{-2 - 16}{-14} \\\\ q &= -1 ~~~~~~~~~~~~,~~ q = \dfrac{9}{7} \end{aligned}
Let's check both solutions to be sure it worked:
q, equals, minus, 1q, equals, start fraction, 9, divided by, 7, end fraction
0=7q2+2q+90=7(1)2+2(1)+90=7(1)2+90=72+90=0\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7(-1)^2+2(-1)+9 \\\\0&=-7(1)-2+9 \\\\0&=-7-2+9\\\\0&=0\end{aligned}0=7q2+2q+90=7(97)2+2(97)+90=7(8149)+(187)+90=(817)+(187)+90=(637)+90=9+90=0\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7\left(\dfrac{9}{7}\right)^2+2\left (\dfrac{9}{7}\right)+9 \\\\0&=-7\left(\dfrac{81}{49}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{81}{7}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{63}{7}\right) +9 \\\\0&=-9 +9 \\\\0&=0\end{aligned}
Yep, both solutions check out.
Want to learn more about the quadratic formula? Check out this video.
Practice
Solve for x.
minus, 4, plus, x, plus, 7, x, squared, equals, 0
Choose 1 answer:
Choose 1 answer:

Want more practice? Check out this exercise.

Want to join the conversation?

  • piceratops ultimate style avatar for user Charlie Williams
    Sal, How does the quadratic formula relate to business and economics?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • female robot ada style avatar for user kaycoach03
    what if the equation doesn't equal zero
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user joeygold2016
    Can u ask Sal to watch an education rap, it's called:

    Bring It Back
    by Balistik (ZT)

    There's a cool little bar about education= the key to success
    (8 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Tiggertast
    can you reccomend other math websites for algebra 1 and 2
    (4 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Drake
    What happens when the discriminant is a negative number? If it is negative would your answer be imaginary?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • hopper happy style avatar for user Palomi Mangesh Kurade
    If a quadratic equation ax^2+bx+c=0 has more than two roots, then it becomes an identity a=b=c=0.

    Can someone please prove this above statement to me?
    - How did the identity part come about? How can a=b=c=0?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      First, in order for an equation to be a quadratic, the "a" (in ax^2+bx+c = 0) can't = 0
      As soon as "a = 0", you no longer have a 2nd degree equation. And, your graph will not be a parabola.

      If a=b=c=0, then the left side of the equation becomes 0. Do the substitution and you'll see.
      Your equation becomes 0=0. This is an equation that is always true, and is called an identity. But, again, it is no longer a quadratic. "a" will never = 0 in a quadratic equation. Quadratics only have 0, 1, or 2 real roots. If they have 0 real roots, then they will have 2 complex roots.
      (4 votes)
  • marcimus purple style avatar for user yaneiri-reyes-gonz
    What happens when a discriminant is a negative number? If it is negative would your answer be imaginary?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user sheilla Muligande
    if the quadratic equation that we are given can be factored, do we factor it down first before plugging it into the quadratic formula?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • orange juice squid orange style avatar for user Mark
    I don't understand how to do this out of solving by graphing, solving by factoring, solving by completing the square, and simplifying square roots, this should be the second easiest behind graphing! Yet somehow, no matter how many times I watch the three videos and read the two articles I never get any of the questions in the Quadratic Formula practice correct, 0/4,0/4,0/4,0/4,0/4, I know I have a bad work ethic but I don't know why I should continue trying to learn this! It's just too hard!
    (1 vote)
    Default Khan Academy avatar avatar for user
    • mr pink green style avatar for user David Severin
      There are several advantages of the quadratic formula including knowing how many roots it has as well as finding answers with irrational numbers (cannot completely simplify the square root). While completing the square also allows you to do this, graphing and factoring cannot accomplish this task. If you want to present one of your attempts step by step, maybe someone can find what you are doing wrong. Several regular issues come up if b is negative. The beginning of the formula, -b, would change the formula to -(-b) or positive b, and if you are squaring this, you have to make sure you are doing (b)^2 inside parentheses rather than b^2 which also messes with apporpriate signs.
      If we have 2x^2 - 4x +5, a=2, b=-4, and c=5. (-b+/- sqrt(b^2-4ac)/(2a) is best done with inside the square root first, so (-4)^2-4(2)(5) = 16-40 which tells us that there are no real solutions. Change c = -5, we end up with (-4)^2-4(2)(-5)=16+40=56. We then have to see if we can simplify sqrt(56). If you try -4^2-4(2)(-5) you get a wrong answer.
      (2 votes)
  • marcimus pink style avatar for user SerafinaFifi
    How do you find the vertex using the quadratic formula?
    (2 votes)
    Default Khan Academy avatar avatar for user