# How do you find the zeros, real and imaginary, of y=-x^2-2x+33 using the quadratic formula?

Jun 25, 2016

$- 1 \pm \sqrt{34}$
$y = - {x}^{2} - 2 x + 33 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 4 + 132 = 136$ --> $d = \pm 2 \sqrt{34}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{2}{-} 2 \pm \frac{2 \sqrt{34}}{-} 2 = - 1 \pm \sqrt{34}$