# How do you find the zeros, real and imaginary, of y= -2x^2-12x-19  using the quadratic formula?

Jul 30, 2016

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

#### Explanation:

$y = - 2 {x}^{2} - 12 x - 19 = 0$
Use the new quadratic formula in graphic form (Socratic Search).
$D = {d}^{2} = {b}^{2} - 4 a c = 144 - 152 = - 8$
Since D < 0, there are no real roots. There are 2 complex roots.
$d = \pm 2 i \sqrt{2}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{12}{-} 4 \pm \frac{2 i \sqrt{2}}{4} = - 3 \pm \frac{i \sqrt{2}}{2}$