# How do I solve this with the quadratic equation?

## $2 {x}^{2} - 8 x - 24$

Feb 23, 2018

$x = 6$ and $x = - 2$

#### Explanation:

This equation is in standard form, $a {x}^{2} + b x + c$. The Quadratic Formula states that the roots to this quadratic are at:

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

In our equation, $a = 2 , b = - 8$ and $c = - 24$. Now, let's plug in:

$x = \frac{8 \pm \sqrt{{\left(- 8\right)}^{2} - 4 \left(2\right) \left(- 24\right)}}{2 \left(2\right)}$

The terms in the radical simplify to:

$x = \frac{8 \pm \sqrt{64 - 8 \left(- 24\right)}}{4}$

$x = \frac{8 \pm \sqrt{64 + 192}}{4}$

$x = \frac{8 \pm \sqrt{256}}{4}$

$x = \frac{8 \pm 16}{4}$

We can factor out a $4$ (because all of the terms are divisible by $4$). We get:

$x = \frac{2 \pm 4}{1}$

Now, let's break up our equations:

$x = 2 + 4$ $\implies x = 6$

$x = 2 - 4$ $\implies x = - 2$

Our zeroes are $x = 6$ and $x = - 2$.

Feb 23, 2018

I have $x = 6$, and $x = - 2$

#### Explanation:

The quadratic equation is as follows:

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

All we have to do is plug the values in and solve.

$a = 2 , b = - 8 , c = - 24$

So, when we plug the numbers in, we get:

(-(-8)+-sqrt((-8)^2-4(2)(-24)))/(2(2)

Now we can start simplifying.

$\frac{8 \pm \sqrt{64 + 192}}{4}$

$\frac{8 \pm \sqrt{256}}{4}$

$\frac{8 \pm 16}{4}$

So our answers will be $x = 6$ and $x = - 2$.