# How do you solve  4x^2 + 2x - 1 = 0 by completing the square?

$x = \frac{\sqrt{5} - 1}{4}$ or $x = - \frac{\sqrt{5} + 1}{4}$
Dividing by $4$ we get
${x}^{2} + \frac{1}{2} x - \frac{1}{4} = 0$
${\left(x + \frac{1}{4}\right)}^{2} = \frac{5}{16}$