# What is the quadratic formula of (2y - 3) (y + 1) = 5?

Oct 25, 2015

Not sure if this is what you were asking for.
$y = \frac{1 \pm \sqrt{65}}{4}$

#### Explanation:

I'm not sure if I understood your question right. Do you want to plug in the values of the quadratic equation into the quadratic formula?

First you need to equate everything to 0. You can start by transferring 5 to the other side.

$\left[1\right] \textcolor{w h i t e}{X X} \left(2 y - 3\right) \left(y + 1\right) = 5$

$\left[2\right] \textcolor{w h i t e}{X X} \left(2 y - 3\right) \left(y + 1\right) - 5 = 0$

Multiply (2y-3) and (y+1).

$\left[3\right] \textcolor{w h i t e}{X X} \left(2 {y}^{2} - y - 3\right) - 5 = 0$

$\left[4\right] \textcolor{w h i t e}{X X} 2 {y}^{2} - y - 8 = 0$

Now just plug the values of $a$, $b$, and $c$ in the quadratic formula.

$a = 2$
$b = - 1$
$c = - 8$

$\left[1\right] \textcolor{w h i t e}{X X} y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\left[2\right] \textcolor{w h i t e}{X X} y = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(2\right) \left(- 8\right)}}{2 \left(2\right)}$

$\left[3\right] \textcolor{w h i t e}{X X} \textcolor{b l u e}{y = \frac{1 \pm \sqrt{65}}{4}}$