# How do you solve (-2/3)x^2 + (-4/3)x + 1 = 0 by completing the square?

May 24, 2015

$0 = - \frac{2}{3} {x}^{2} - \frac{4}{3} x + 1 = - \frac{2}{3} \left({x}^{2} + 2 x - \frac{3}{2}\right)$

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

$= - \frac{2}{3} \left({\left(x + 1\right)}^{2} - \frac{5}{2}\right)$

Divide both sides by $- \frac{2}{3}$ to get:

$0 = {\left(x + 1\right)}^{2} - \frac{5}{2}$

Add $\frac{5}{2}$ to both sides to get:

${\left(x + 1\right)}^{2} = \frac{5}{2}$

Then

$x + 1 = \pm \sqrt{\frac{5}{2}}$

Subtract $1$ from both sides to get:

$x = - 1 \pm \sqrt{\frac{5}{2}} = - 1 \pm \frac{\sqrt{10}}{2}$