# How do you find the roots, real and imaginary, of y=-3x^2 + -3x -4  using the quadratic formula?

Jul 12, 2016

$x = - \frac{1}{2} \pm \frac{i \sqrt{39}}{6}$
$y = - 3 {x}^{2} - 3 x - 4 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 9 - 48 = - 39$ < 0 --> $d = \pm i \sqrt{39}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{3}{-} 6 \pm i \frac{\sqrt{39}}{6} = - \frac{1}{2} \pm \frac{i \sqrt{39}}{6}$