Help w/ algebra questions

1. what is the solution set of the equation x^2-5x=0
2. the perimeter of a square is 52. express the length of a diagonal of a square in simplest radical form.

x(x-5)=0
x=0, 5

4x=52
x=13

13^2+13^2=c^2
338=c^2
√338=c^2
√169*√2=c
13√2=c

1.
x^2 - 5x = 0
x(x - 5) = 0
x = [0, 5]


2.
Let s = length of 1 side
4s = 52
s = 13

Let d = length of diagonal
13^2 + 13^2 = d^2
169 + 169 = d^2
338 = d^2
d = sqrt(338)
d = sqrt(13^2 * 2)
d = 13 * sqrt(2)

1)
x = 0 || x = 5

2)

Sqrt[13^2 + 13^2]

13 Sqrt[2]