# How do you solve  y^2 - 3y - 14 = 0 by completing the square?

${\left(y - \left(\frac{3}{2}\right)\right)}^{2} = {y}^{2} - 3 y + \frac{9}{4}$,
so ${y}^{2} - 3 y - 14 = {\left(y - \frac{3}{2}\right)}^{2} - \left(14 + \frac{9}{4}\right)$
$= {\left(y - \frac{3}{2}\right)}^{2} - \frac{65}{4}$.
And, since we are given that ${y}^{2} - 3 y - 14 = 0$
${\left(y - \frac{3}{2}\right)}^{2} = \frac{65}{4}$
Hence $y = \pm \sqrt{\frac{65}{4}} + \frac{3}{2} = \pm \frac{\sqrt{65}}{2} + \frac{3}{2}$.