# How do you find the zeros, real and imaginary, of y=-x^2-6x+11 using the quadratic formula?

May 8, 2018

$\left(- 3 + 2 \sqrt{5} , 0\right) \mathmr{and} \left(- 3 - 2 \sqrt{5} , 0\right)$

#### Explanation:

$y = - {x}^{2} - 6 x + 11$

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:
$\textcolor{red}{a = - 1}$

$\textcolor{m a \ge n t a}{b = - 6}$

$\textcolor{b l u e}{c = 11}$

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:

$x = \frac{- \left(\textcolor{m a \ge n t a}{- 6}\right) \pm \sqrt{{\left(\textcolor{m a \ge n t a}{- 6}\right)}^{2} - 4 \left(\textcolor{red}{- 1}\right) \left(\textcolor{b l u e}{11}\right)}}{2 \left(\textcolor{red}{- 1}\right)}$

Simplify:
$x = \frac{6 \pm \sqrt{36 + 44}}{-} 2$

$x = \frac{6 \pm \sqrt{80}}{-} 2$

$x = \frac{6 \pm 4 \sqrt{5}}{-} 2$

$x = - 3 \pm 2 \sqrt{5}$

This is the same thing as:
$x = - 3 + 2 \sqrt{5}$ and $x = - 3 - 2 \sqrt{5}$
because $\pm$ means "plus or minus."

This means the zeros are at:
$\left(- 3 + 2 \sqrt{5} , 0\right) \mathmr{and} \left(- 3 - 2 \sqrt{5} , 0\right)$

Hope this helps!