# How do you find the zeros, real and imaginary, of y= -x^2-5x-14 using the quadratic formula?

Nov 22, 2015

Two imaginary roots $\left(- \frac{5 \pm \sqrt{31} i}{2}\right)$

#### Explanation:

For a quadratic equation in the general form
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
gives us the zeros (or "roots") of the equation.

For the specific equation
$\textcolor{w h i t e}{\text{XXX}} y = - {x}^{2} - 5 x - 14$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}} a = \left(- 1\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}} b = \left(- 5\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}} c = \left(- 14\right)$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(- 1\right) \left(- 14\right)}}{2 \left(- 1\right)}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{5 \pm \sqrt{25 - 56}}{- 2}$
$\textcolor{w h i t e}{\text{XXX}} = - \frac{5 \pm \sqrt{31} i}{2}$