# How do you solve (x-5)^2=4?

Jul 16, 2016

We must find which numbers $t$ satisfy ${t}^{2} = 4$

#### Explanation:

If ${t}^{2} = 4$, then either $t = 2$ or $t = - 2$.

But then we have from the given equation that either $\left(x - 5\right) = 2$ or $\left(x - 5\right) = - 2$. Thus we have either:

$x = 5 + 2$, or $x = 5 - 2$. That is either:

$x = 7$, or $x = 3$, which are the solutions we are looking for.