# How do you find the zeros, real and imaginary, of y=x^2+32x-4 using the quadratic formula?

Apr 29, 2018

$x = - 16 \pm 2 \sqrt{65}$

#### Explanation:

$y = {x}^{2} + 32 x - 4$

To use the quadratic formula to find the zeroes, we need to make sure the equation is written in the form $\textcolor{red}{a} {x}^{2} + \textcolor{m a \ge n t a}{b} x + \textcolor{b l u e}{c} = 0$, which this equation is.

So we know that:
$\textcolor{red}{a = 1}$

$\textcolor{m a \ge n t a}{b = 32}$

$\textcolor{b l u e}{c = - 4}$

The quadratic formula is $x = \frac{- \textcolor{m a \ge n t a}{b} \pm \sqrt{{\textcolor{m a \ge n t a}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{b l u e}{c}}}{2 \textcolor{red}{a}}$.

Now we can plug in the values for $\textcolor{red}{a}$, $\textcolor{m a \ge n t a}{b}$, and $\textcolor{b l u e}{c}$ into the quadratic formula:

$x = \frac{- \textcolor{m a \ge n t a}{32} \pm \sqrt{{\left(\textcolor{m a \ge n t a}{32}\right)}^{2} - 4 \left(\textcolor{red}{1}\right) \left(\textcolor{b l u e}{- 4}\right)}}{2 \left(\textcolor{red}{1}\right)}$

Simplify:
$x = \frac{- 32 \pm \sqrt{1024 + 16}}{2}$

$x = \frac{- 32 \pm \sqrt{1040}}{2}$

$x = \frac{- 32 \pm 4 \sqrt{65}}{2}$

$x = - 16 \pm 2 \sqrt{65}$

This is the same thing as:
$x = - 16 + 2 \sqrt{65}$ and $x = - 16 - 2 \sqrt{65}$
because $\pm$ means "plus or minus."

Hope this helps!