# How do you find the zeros, real and imaginary, of y= -x^2-12x-8  using the quadratic formula?

Jan 31, 2017

x=-6+2sqrt7; or $x = - 6 - 2 \sqrt{7}$

#### Explanation:

When using the quadratic formula, the zeros are the two values for $x$. When solving quadratic equations, the equation must be equal to $0$.

Your equation: $y = - {x}^{2} - 12 x - 8$

Set $y$ equal to zero.

$0 = - {x}^{2} - 12 x - 8$

Multiply by $- 1$ to make $a$ positive.

${x}^{2} + 12 x \pm 8$

$- {x}^{2} - 12 x - 8$ is a quadratic equation in the form of "color(red)(a)^2+color(blue)bx+color(green)(c), where the coefficients are "color(red)(a)=(1), $\textcolor{b l u e}{b} = 12$, $\textcolor{g r e e n}{c} = 8$.

$x = \frac{\textcolor{b l u e}{- b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$

Substitute the known values into the equation.

x=("-(color(blue)(12))+-sqrt(color(blue)(12)^2-4(color(red)(1))(color(green)(8))))/((2*color(red)(1)))

$x = \frac{- 12 \pm \sqrt{144 - \left(32\right)}}{- 2}$

$= \frac{- 12 \pm \sqrt{112}}{- 2}$

Factor $\sqrt{112}$.

$\sqrt{112} = \sqrt{16 \times 7} = 4 \sqrt{7}$

$x = \frac{- 12 \pm 4 \sqrt{7}}{2}$

Simplifyby factoring out $2$ in $- 12 , \pm 4 , \mathmr{and} 2$.

$x = - 6 \pm 2 \sqrt{7}$

x=-6+2sqrt7; or $x = - 6 - 2 \sqrt{7}$