How can you solve this

I am bit confused in this equation. I knew this before but now, I forgot it. Here is the equation:
x = √(2+√(2+√(2+...))); solve for x (The answer in here is 2)
~This one is continuous radical equation.What is the solution for this?

Another one is:
√(y+(√y+(√y+...))) = 5; Solve for y (The answer is 20)
~What is the solution for this?

x=√(2+√(2+√(2+...))).
x=√(2+x) [since x=√(2+√(2+√(2+...))) until infinity]
Squaring on both sides.
x^2=2+x
x^2-2-x=0
x^2-x-2=0
x^2-2x+x-2=0
(x^2-2x)+(x-2)=0
x(x-2)+1(x-2)=0
(x-2)(x+1)=0
So x=2 or -1

In the next question,
5=√(y+(√y+(√y+...)))
5=√(y+5) [For the same reason as in the first problem]
Squaring on both sides,
25=y+5
y+5=25
y=25-5
=20

No, I can't.