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

Mar 18, 2016

$x = - 3 \pm \sqrt{11}$

#### Explanation:

$D = {d}^{2} = {b}^{2} - 4 a c = 36 + 8 = 44$ --> $d = \pm 2 \sqrt{11}$
There are 2 real roots:
$x = \frac{6}{-} 2 \pm 2 \frac{\sqrt{11}}{-} 2 = - 3 \pm \sqrt{11}$