# How do you solve (x-5)^2=9?

Only two numbers are such that raised to the square $= 9$
If a number to the square $= 9$, it is either $3$ or $- 3$. So, either:
$x - 5 = 3$, or $x - 5 = - 3$. Then, it is either $x = 3 + 5$ or $x = - 3 + 5$
The two solutions are then $x = 8$ and $x = 2$