# How do you solve 3x^2 - 12x + 5=0 by completing the square?

May 1, 2015

Write it as:
${x}^{2} - \frac{12}{3} x = - \frac{5}{3}$
${x}^{2} - 4 x = - \frac{5}{3}$ add and subtract 4
${x}^{2} - 4 x + 4 - 4 = - \frac{5}{3}$
${\left(x - 2\right)}^{2} - 4 = - \frac{5}{3}$
${\left(x - 2\right)}^{2} = 4 - \frac{5}{3}$
${\left(x - 2\right)}^{2} = \frac{7}{3}$
$x - 2 = \pm \sqrt{\frac{7}{3}}$
Finally:
${x}_{1 , 2} = 2 \pm \sqrt{\frac{7}{3}}$