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

Jul 6, 2015

Add 14 to both sides, then add half the square of the coefficient of $y$ to both sides, then solve.

#### Explanation:

1) Add 14 to both sides of the equation

${y}^{2} - 3 y = 14$

2) Add half the square of $y$'s coefficient to both sides

${y}^{2} - 3 y + {\left(- \frac{3}{2}\right)}^{2} = 14 + {\left(- \frac{3}{2}\right)}^{2}$

3) Now the left side is a perfect square; remember to simplify the right hand side too

${\left(y - \frac{3}{2}\right)}^{2} = \frac{65}{4}$

4) Take $\pm$ the square root of both sides; remember the left side is a square, so it remains positive (i.e., it's not $\pm$)

$y - \frac{3}{2} = \pm \sqrt{\frac{65}{4}}$

5) Add the remaining fraction on the left to both sides and simplify your roots

$y = \frac{3}{2} \pm \frac{\sqrt{65}}{2}$

The solution set is: $\left[\frac{3}{2} - \frac{\sqrt{65}}{2} , \frac{3}{2} + \frac{\sqrt{65}}{2}\right]$