# How do you use the quadratic formula to find both solutions to the quadratic equation (2y - 3) (y + 1) = 5?

Jul 6, 2015

Multiply out and rearrange into the form $a {y}^{2} + b y + c = 0$ then use the quadratic formula to find:

$y = \frac{1 \pm \sqrt{65}}{4}$

#### Explanation:

$5 = \left(2 y - 3\right) \left(y + 1\right) = 2 {y}^{2} - y - 3$

Subtract $5$ from both sides to get:

$2 {y}^{2} - y - 8 = 0$

This is of the form $a {y}^{2} + b y + c = 0$

with $a = 2$, $b = - 1$ and $c = - 8$

This has solutions given by the quadratic formula:

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - \left(4 \times 2 \times - 8\right)}}{2 \cdot 2}$

$= \frac{1 \pm \sqrt{65}}{4}$