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

May 27, 2018

$x = - \frac{1}{3} + \frac{2}{3} \sqrt{2} i$ or $x = - \frac{1}{3} - \frac{2}{3} \sqrt{2} i$

Explanation:

Given
$3 {x}^{2} - 2 x + 3 = 0$
dividing by $3$
${x}^{2} - \frac{2}{3} x + 1 = 0$
Using the quadratic formula
${x}_{1 , 2} = - \frac{1}{3} \pm \sqrt{\frac{1}{9} - \frac{9}{9}}$
this is
${x}_{1 , 2} = - \frac{1}{3} \pm \setminus \sqrt{\frac{8}{9}} i$
simplifying
${x}_{1 , 2} = - \frac{1}{3} \setminus \pm \frac{2}{3} \setminus \sqrt{2} i$