So you need the function H(x) to be able to take in G(x) = (x-3)^1/2 +2 and put out x. Basically, H(x) has to undo whatever math is embedded in G(x).
Since you start with G(x) = (x-3)^1/2 + 2 and you want to work outside-in the first thing you want to do is subtract 2. Next you want to cancel out the square root by squaring it, so the new function is (x-2)^2. Now all that's left is x-3, so you add 3 to get your new function, (x-2)^2 + 3