# How do you find the roots, real and imaginary, of y=3x^2 -11x + 4 using the quadratic formula?

May 10, 2016

$\frac{11 \pm \sqrt{73}}{6}$
$y = 3 {x}^{2} - 11 x + 4 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 121 - 48 = 73$ --> $d = \pm \sqrt{73}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{11}{6} \pm \frac{\sqrt{73}}{6} = \frac{11 \pm \sqrt{73}}{6}$