# How do you solve x^2+3x+1=0?

May 29, 2015

We can solve this by completing the square ...

$0 = {x}^{2} + 3 x + 1$

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

$= {\left(x + \frac{3}{2}\right)}^{2} - \frac{5}{4}$

Add $\frac{5}{4}$ to both ends to get:

${\left(x + \frac{3}{2}\right)}^{2} = \frac{5}{4}$

So

$x + \frac{3}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{\sqrt{4}} = \pm \frac{\sqrt{5}}{2}$

Subtract $\frac{3}{2}$ from both ends to get:

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