# How do you solve using the quadratic formula (3 - y)(y + 4) = 3y - 5?

##### 1 Answer
Apr 29, 2015

First convert the given equation into a standard quadratic form by simplification
$\left(3 - y\right) \left(y + 4\right) = 3 y - 5$

$3 y + 12 - {y}^{2} - 4 y = 3 y - 5$

$- {y}^{2} - y + 12 = 3 y - 5$

$= - {y}^{2} - 4 y + 17 = 0$

${y}^{2} + 4 y - 17 = 0$

Now apply the quadratic formula for roots
$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$

$= \frac{- 4 \pm \sqrt{16 + 68}}{2}$

$= 2 \pm \sqrt{21}$