# How do you solve 3x^2 + 5x = -x + 4 by completing the square?

##### 1 Answer
May 19, 2015

y = 3x^2 + 6x - 4 = 3(x^2 + 2x - 4/3) = 0

[x^2 + 2x - 4/3 = x^2 + 2x + 1 - 1 - 4/3 = 0

$y = {\left(x + 1\right)}^{2} - \frac{7}{3} = 0$

${\left(x + 1\right)}^{2} = \frac{7}{3}$ -> $\left(x + 1\right) = \pm \frac{\sqrt{7}}{\sqrt{3}}$

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