# How do you find the roots, real and imaginary, of y= 4x^2-3x-6  using the quadratic formula?

$x = \frac{3 \pm \sqrt{105}}{8}$
$y = 4 {x}^{2} - 3 x - 6 = 0$.
$D = {d}^{2} = {b}^{2} - 4 a c = 9 + 96 = 105$ --> $d = \pm \sqrt{105}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{3}{8} \pm \frac{\sqrt{105}}{8} = \frac{3 + \sqrt{105}}{8}$