# How do you solve 3x^2 + 4x + 3 = 0 by completing the square?

Sep 10, 2017

#### Answer:

$x = \pm \frac{i \sqrt{- 5}}{3} - \frac{2}{3}$

#### Explanation:

Given -

$3 {x}^{2} + 4 x + 3 = 0$
Take the constant term to right

$3 {x}^{2} + 4 x = - 3$

Divide each term on both sides by the coefficient of ${x}^{2}$

$\frac{3 {x}^{2}}{3} + \frac{4 x}{3} = \frac{- 3}{3}$

${x}^{2} + \frac{4}{3} x = - 1$

Divide the coefficient $x$, square it and add it to both sides

${x}^{2} + \frac{2}{3} x + \frac{4}{9} = - 1 + \frac{4}{9} = \frac{- 9 + 4}{9} = - \frac{5}{9}$

${\left(x + \frac{2}{3}\right)}^{2} = - \frac{5}{9}$
Take square root on both sides

$\left(x + \frac{2}{3}\right) = \sqrt{- \frac{5}{9}} = \frac{\sqrt{- 5}}{3}$

$x = \pm \frac{i \sqrt{- 5}}{3} - \frac{2}{3}$