This equation has no solutions
Move the square root of y to the right hand sides:
[tex] \sqrt{y-3} = \sqrt{y}-3 [/tex]
To eliminate the roots, let's square both sides. But there is a crucial observation in this passage: you are allowed to deduce [tex] a=b \implies a^2=b^2 [/tex] only if both [tex] a [/tex] and [tex] b [/tex] are positive. Otherwise, you risk contraddictions like [tex] 2 = -2 \implies 4 = 4 [/tex].
Since the left hand side is always positive when defined (it's a square root), we need to ask [tex] \sqrt{y}-3 > 0 \iff \sqrt{y} > 3 \iff x>9 [/tex]. So, we're only accept answers if they're greater than nine.
Now we can square both sides:
[tex] y-3 = y- 6\sqrt{y} + 9 [/tex]
Simplify y's and bring 9 to left hand side:
[tex] -12 = -6\sqrt{y} [/tex]
Divide both sides by -6:
[tex] 2 = \sqrt{y} \implies y = 4 [/tex]
So, the solution is not acceptable. In fact, it would lead to
[tex] \sqrt{4-3} - \sqrt{4} = 1-2 \neq -3 [/tex]
