The Quadratic Formula


Sometimes you won’t be able to solve a quadratic equation by factoring because not everything in the world works according to nice, even numbers. In this case, the only algebraic option you have without resorting to graphing is the use of the quadratic formula. Assuming a quadratic equation in the form ax² + bx + c = 0, the quadratic formula tells us the following:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} 

Note that this gives us both values of x by the use of the plus and minus sign. So let’s use an example. On another page on this site, we solved 6x² + 11x – 10 = 0 using factoring and found that the answers were x = 2/3 and x = -5/2. Let’s use the quadratic formula to see what we get. Here, a = 6, b = 11, and c = -10.

x = \frac{-(11) \pm \sqrt{(11)^2 - 4(6)(-10)}}{2(6)} 
x = \frac{-11 \pm \sqrt{121 + 240}}{12} 
x = \frac{-11 \pm \sqrt{361}}{12} 
x = \frac{-11 \pm 19}{12} 

This gives us two results, one when we use addition and the other when we use subtraction. Let’s start with addition:

x = \frac{-11 + 19}{12} 
x = \frac{8}{12} 
x = \frac{2}{3} 

Now for the result with subtraction:

x = \frac{-11 - 19}{12} 
x = -\frac{30}{12} 
x = -\frac{5}{2} 

And we see pretty clearly that we get the same answer.

Note that it’s possible to get answers that are complex numbers. This happens in the case of quadratic equations like x² + 1 = 0 where there are no real solutions. Algebraically, if we solve for x in this equation we get:

x² + 1 = 0
x² = -1
x = i and x = -i

Like always, we can substitute our answers back into the original equation to check our answer. Let’s start with i:

x² + 1 = 0
i² + 1 = 0
-1 + 1 = 0
0 = 0

Now with -i:

x² + 1 = 0
(-i)² + 1 = 0
-1 + 1 = 0
0 = 0

And there you have it. The quadratic formula can be used to solve any quadratic equation of one variable, and is the only method you have to work out the answers for a quadratic equation with just a pencil and paper for the vast majority of equations. Later on, you’ll be learning how to graph quadratic equations by hand, and how you could solve them graphically.