# How do you find the zeros, real and imaginary, of y=-4x^2-4x-16 using the quadratic formula?

Jul 8, 2017

$- \frac{1 \pm i \sqrt{15}}{2}$

#### Explanation:

The standard form of a quadratic equation is $y = a {x}^{2} + b x + c$. Based on this, you know that $a = - 4$, $b = - 4$, and $c = - 16$.

The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. Substitute the values above into the formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(- 4\right) \left(- 16\right)}}{2 \left(- 4\right)}$

$x = \frac{4 \pm \sqrt{- 240}}{- 8}$

$x = \frac{4 \pm 4 i \sqrt{15}}{- 8}$

$x = \frac{1 \pm i \sqrt{15}}{- 2} \to - \frac{1 \pm i \sqrt{15}}{2}$

There are no real zeros.