# How do you find the zeros, real and imaginary, of y= 3x^2-6x+14 using the quadratic formula?

Jul 12, 2016

$1 \pm \frac{i \sqrt{33}}{3}$
$y = 3 {x}^{2} - 6 x + 14 = 0$
$D = {d}^{2} = 36 - 168 = - 132 < 0$
$d = \pm 2 i \sqrt{33}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{6}{6} \pm i \frac{\sqrt{33}}{3} = 1 \pm i \frac{\sqrt{33}}{3}$