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

Mar 28, 2016

$x = - 3$ or $x = - 1$

#### Explanation:

Add $1$ to both sides of the equation to get:

${x}^{2} + 4 x + 4 = 1$

The left hand side is now a perfect square:

${x}^{2} + 4 x + 4 = {\left(x + 2\right)}^{2}$

So we have:

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

Hence:

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

Subtract $2$ from both ends to get:

$x = - 2 \pm 1$

That is:

$x = - 3$ or $x = - 1$