How do you find the zeros, real and imaginary, of y= -x^2-6x-4 using the quadratic formula?

Jan 14, 2016

$x = - 3 \pm \sqrt{5}$

Explanation:

For any quadratic equation $y = a {x}^{2} + b x + c$, the zeroes are found through the formula:

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

$a = - 1$
$b = - 6$
$c = - 4$

Plug these into the quadratic formula.

$x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - \left(4 \times - 1 \times - 4\right)}}{2 \times - 1}$

Which simplifies to be

$x = \frac{6 \pm \sqrt{36 - 16}}{- 2}$

$x = \frac{- 6 \pm \sqrt{20}}{2}$

$x = \frac{- 6 \pm 2 \sqrt{5}}{2}$

$x = - 3 \pm \sqrt{5}$