# How do you solve x^2+3x+1=0 by completing the square?

${x}^{2} + 3 x + 1 = {\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4} + 1 = {\left(x + \frac{3}{2}\right)}^{2} - \frac{5}{4}$
Now remember the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
In our problem we write (because is the same) ${\left(x + \frac{3}{2}\right)}^{2} - {\sqrt{5}}^{2} / {2}^{2}$ applying identity with $a = x + \frac{3}{2}$ and $b = \frac{\sqrt{5}}{2}$ and we have
$\left(x + \frac{3}{2} + \frac{\sqrt{5}}{2}\right) \left(x + \frac{3}{2} - \frac{\sqrt{5}}{2}\right)$ or the equivalent
$\left(x + \frac{3 + \sqrt{5}}{2}\right) \left(x + \frac{3 - \sqrt{5}}{2}\right)$