This lesson right here is probably the one most feared by Algebra students! The quadratic formula looks like a huge, scary expression, but really, it’s not that bad.

The quadratic formula is used to find the roots of a quadratic equation. We used the discriminant to find out how many roots there were, but the quadratic equation will actually tell us what they are. The roots are the points where the equation equals zero, which is the same as the points where the graph hits the x-axis. Remember, there can be two real roots, one real root, or no real roots.

So, are you ready? Let’s try one:

\(x^2 + 5x = -1\)

We are going to need to find our a, b, & c which means we need the equation to equal zero. So let’s add 1 to each side.

\(\array{ x^2 + 5x  =&  -1 \cr +1 & +1}\)

\(x^2 + 5x + 1 = 0\)

Remember the standard form is \(ax^2 + bx + c\) so we have:

\(x^2 + 5x + 1 = 0\)

\(a = 1, b = 5, c = 1\)

Now, here comes the quadratic formula. It’s a lot of plugging in and simplifying using your a, b, and c.

Quadratic Formula

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

The "\(\pm\)" symbol means “plus or minus” which means this formula is actually two in one! We will have to split it later on. Let’s start plugging in our numbers.

\(a = 1, b = 5, c = 1\)

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

\(x=\Large \frac{-5 \pm \sqrt{(5)^2 - 4(1)(1)}}{2(1)}\)

Now let’s simplify.

\(x=\Large \frac{-5 \pm \sqrt{(5)^2 - 4(1)(1)}}{2(1)}\)

\(x=\Large \frac{-5 \pm \sqrt{25 - 4}}{2}\)

\(x=\Large \frac{-5 \pm \sqrt{21}}{2}\)

Sometimes, you will be allowed to keep your answer in this form. If a decimal form is asked for, then we will have to split the "\(\pm\)" and estimate the \(\sqrt{21}\).

Whew! So, that wasn’t too tough but definitely is a lot of work! Let’s practice and try one more:

\(-2p^2 + 16 = 8 - p^2\)

We want it to equal zero so subtract the 8 and add the \(p^2\) to both sides.

\(\array{ -2p^2 + 16  =&8 - p^2 \cr -8 & -8}\)

\(\array{ -2p^2 + 8  =&- p^2 \cr +p^2 & +p^2}\)

\(-p^2 + 8 = 0\)

“b” is missing! That just means it equals zero.

a = -1, b = 0, c = 8

\(p=\Large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(p=\Large \frac{-(0) \pm \sqrt{(0)^2 - 4(-1)(8)}}{2(-1)}\)

\(p=\Large \frac{-0 \pm \sqrt{0 - (-32)}}{-2}\)

\(p=\Large \frac{\pm \sqrt{32}}{-2}\)

And if it needs to be estimated, then split them!

Done! My suggestion for these problems is to take your time and keep your work neat. They are long problems and one little mistake could make a big mess! So double-check everything and the quadratic formula won’t seem so bad after all.

Bellow you can download some free math worksheets and practice.