# How do you find the roots, real and imaginary, of y= 12x^2+3x-6  using the quadratic formula?

Feb 15, 2016

$x = - \frac{1}{8} \pm \frac{\sqrt{33}}{8}$
$y = 12 {x}^{2} + 3 x - 6 = 4 {x}^{2} + x - 2 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 1 + 32 = 33$ --> $d = \pm \sqrt{33}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{1}{8} \pm \frac{\sqrt{33}}{8}$