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

Jul 1, 2017

$x = - 1 \pm \sqrt{\frac{5}{3}}$

#### Explanation:

We have:

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

The standard steps to complete the square on quadratic expression are as follows:

Step 1 - Factor out the quadratic coefficient, thus:

$3 {x}^{2} + 6 x - 2 = 3 \left\{{x}^{2} + 2 x - \frac{2}{3}\right\}$

Step 2 - Factor $\frac{1}{2}$ of the $x$ coefficient to form a perfect square, and subtract its square, and simplifiy, thus:

$3 {x}^{2} + 6 x - 2 = 3 \left\{{\left(x + \frac{2}{2}\right)}^{2} - {\left(\frac{2}{2}\right)}^{2} - \frac{2}{3}\right\}$
$\text{ } = 3 \left\{{\left(x + 1\right)}^{2} - 1 - \frac{2}{3}\right\}$
$\text{ } = 3 \left\{{\left(x + 1\right)}^{2} - \frac{5}{3}\right\}$

So now returning to the quadratic equation , we have;

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

$\therefore 3 \left\{{\left(x + 1\right)}^{2} - \frac{5}{3}\right\} = 0$
$\therefore {\left(x + 1\right)}^{2} - \frac{5}{3} = 0$
$\therefore {\left(x + 1\right)}^{2} = \frac{5}{3}$
$\therefore x + 1 = \pm \sqrt{\frac{5}{3}}$
$\therefore x = - 1 \pm \sqrt{\frac{5}{3}}$