# How do you solve 0=3x^2-11x+6 by completing the square?

Jun 5, 2015

$0 = 3 {x}^{2} - 11 x + 6$

$= 3 \left({x}^{2} - \frac{11}{3} x + 2\right)$

$= 3 \left({x}^{2} - \frac{11}{3} x + \frac{121}{36} - \frac{121}{36} + 2\right)$

$= 3 \left({\left(x - \frac{11}{6}\right)}^{2} - \frac{121}{36} + \frac{72}{36}\right)$

$= 3 \left({\left(x - \frac{11}{6}\right)}^{2} - \frac{49}{36}\right)$

Divide both sides by $3$ to get:

${\left(x - \frac{11}{6}\right)}^{2} - \frac{49}{36} = 0$

Add $\frac{49}{36}$ to both sides to get:

${\left(x - \frac{11}{6}\right)}^{2} = \frac{49}{36}$

Take square root of both sides to get:

$x - \frac{11}{6} = \pm \sqrt{\frac{49}{36}} = \pm \frac{7}{6}$

Add $\frac{11}{6}$ to both sides to get:

$x = \frac{11}{6} \pm \frac{7}{6} = \frac{11 \pm 7}{6}$

That is $x = 3$ or $x = \frac{2}{3}$