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

Aug 8, 2017

$x = = 1 + \sqrt{3} i$
$x = - 1 - \sqrt{3} i$

#### Explanation:

Identify the values of a,b & c and substitute in using the quadratic formula: $x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = - 1$

$b = - 2$

$c = - 4$

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

$x = \frac{2 \setminus \pm \sqrt{4 - 16}}{- 2}$

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

$x = \frac{2 \setminus \pm 2 \sqrt{3} i}{- 2}$

$x = - 1 \setminus \pm \sqrt{3} i$

$x = 1 + \sqrt{3} i$, $x = - 1 - \sqrt{3} i \leftarrow$ Final solutions