Hard radical problem help

problem: √ √2y - √y-1 = 1
Step 1: (√ √2y - √y-1)^2 = (1)^2
Step 3: √2y - √y-1 = 1

I cant seem to figure out the rest of the problem can you please help me

Starting from where you left off:

√(2y) - √(y - 1) = 1 (subtract √(2y) from both sides)
-√(y - 1) = -√(2y) + 1 (multiply or divide both sides by -1)
√(y - 1) = √(2y) + 1 (square both sides)
y - 1 = (√(2y) + 1)(√(2y) + 1) (FOIL)
y - 1 = 2y + 2√(2y) + 1 (subtract 2y and 1 from both sides)
-y - 2 = 2√(2y) (square both sides)
y^2 + 4y + 4 = 4(2y) (multiply)
y^2 + 4y + 4 = 8y (subtract 8y from both sides)
y^2 - 4y + 4 = 0 (factor)
(y - 2)(y - 2) = 0 (zero-product principle)
y - 2 = 0 (add 2 to both sides)
y = 2 <===ANSWER

After this, I just squared both sides to get:
2y - (y - 1) = 1

Then I distributed:

2y - y + 1 = 1

Then I simplified:

y = 0


I'm not a genius at radicals, but that's how I'd do the problems.

simplify and repeat step 1
√2y - √y-1 = 1
√2y - √y = 2
(√2y - √y)^2 = (2)^2
etc...