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

Feb 14, 2016

${x}_{1} = - 4$
${x}_{2} = 2$

#### Explanation:

$y = a {x}^{2} + b x + c$

We can find its zeros using the Quadratic Formula:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$y = - {x}^{2} - 2 x + 8$

$\implies x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot \left(- 1\right) \cdot \left(8\right)}}{2 \left(- 1\right)}$

$\implies x = \frac{2 \pm \sqrt{4 - \left(- 32\right)}}{-} 2$

$\implies x = \frac{2 \pm \sqrt{36}}{-} 2$

$\implies x = \frac{2 \pm 6}{-} 2$

$\implies {x}_{1} = \frac{8}{-} 2 = - 4$

$\implies {x}_{2} = - \frac{4}{-} 2 = 2$