# How do you solve by completing the square for 3x^2 = 6x + 4?

May 23, 2015

$f \left(x\right) = 3 {x}^{2} - 6 x - 4$ = 0

f(x) = 3(x^2 - 2x - 4/3) = 3(x^2 - 2x + 1) - 3 - 12/3

(x - 1)^2 = 21/9 =

$\left(x - 1\right) = \pm \frac{\sqrt{21}}{3}$ ->

$\left(x - 1\right) = \frac{\sqrt{21}}{3} \to x = 1 + \frac{\sqrt{21}}{3}$

$\left(x - 1\right) = - \frac{\sqrt{21}}{3} \to x = 1 - \frac{\sqrt{21}}{3}$