# How do you solve y=(x+2)^2-5?

Jun 27, 2015

x= -2$\pm \sqrt{5}$

#### Explanation:

Solving the given equation would imply finding the roots of the quadratic on the right side. This can be done by making y=0 and solving for x

${\left(x + 2\right)}^{2} - 5 = 0$

${\left(x + 2\right)}^{2} = 5$

x+2= $\pm \sqrt{5}$

x= -2$\pm \sqrt{5}$