If you keep track, there is little risk.
2x - 3 >= 0 leads to x >= 3/2
x + 2 >= 0 leads to x >= -2
Therefore, if we promise to keep x >= 3/2, we generally can do as we wish.
The second methodology would be to check final potential answers in ORIGINAL equations if they exist.
Please don't just square in the present condition. First, isolate one of the radicals.
There are other failsafes that we can track along the way.
\(\sqrt{2x-3} = 1 + \sqrt{x+2}\). Everything is positive, here. There should be little or no risk in the squaring.
\(2x - 3 = 1 + 2\sqrt{x+2} + (x+2)\)
Simplify, Isolate, and square it again. ALWAYS keep your eyes on the Domain issues.