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

$1 \pm \frac{\sqrt{3}}{3}$
$D = {d}^{2} = {b}^{2} - 4 a c = 36 - 24 = 12 - \to d = \pm 2 \sqrt{3}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{6}{6} \pm \frac{2 \sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$