(x-5)^2=64 Solve the following equation

please help me :(

Use the square root property to square root both sides, undoing the squaring of (x-5). Once you square root both sides you should be left with "(x-5) = 8."

Next, use the Addition Property of Equality to add 5 to both sides of the equation. After adding 5, you should be left with "x = 13."

Answer: x = 13

The trick here is to see that 64 is a perfect square

the square root of 64 = +/- 8

so set up two equations
(x - 5) = - 8 or (x - 5) = 8 and 5 to both sides of each equation

x = - 3 and x = 13

Take the square root of (x-5)^2 and 64
Here, you have x-5 = 8
Add five to the other side
x=13

take the sqaure roots of both sides

x-5=8________________ x-5=-8
x=13_________________ x=-3