# How do you solve 2y² + 2y = 1 using the quadratic formula?

Jun 23, 2016

$y = - \frac{1}{2} \pm \frac{\sqrt{3}}{2}$
(see below for use of quadratic formula)

#### Explanation:

$2 {y}^{2} + 2 y = 1$ is equivalent to $\textcolor{red}{2} {y}^{2} + \textcolor{b l u e}{2} y \textcolor{g r e e n}{- 1}$

$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {y}^{2} + \textcolor{b l u e}{b} y + \textcolor{g r e e n}{c} = 0$
color(white)("XXX")y=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4(color(red)(a))(color(green)(c))))/(2(color(red)(a))
color(white)("XXX")y=(-color(blue)(2)+-sqrt(color(blue)(2)^2-4(color(red)(2))(color(green)(-1))))/(2(color(red)(2))
$\textcolor{w h i t e}{\text{XXX}} y = \frac{- 2 \pm \sqrt{12}}{4} = - \frac{1}{2} \pm \frac{\sqrt{3}}{2}$