Raising a number to the 1/2 power is the same as taking the square root. The answer will be positive (the principle root.)
There are two solutions to y^2=4; the solutions are 2 and -2. Thus if you take the square root of both sides (or raise both sides to the 1/2 power) you must put +/- on the right hand side.
When raising both sides to an even power as in this problem, you should either determine the domain (as you have done) or check for extraneous solutions;i.e. substitute the possible solutions into the original equation to check if they work.
Let the given problem be named as (1).
Let us first define the domain of y (This step is important. You might get away in some questions without finding the domain but this is no a long term solutions. Good exams with check your knowledge of domain). The quantity inside the square root sign must always be non-negative. Thus `63-3y gt= 0` .
Thus `y lt= 21` .......... (2)
Also, the value that comes from a square root is always positive. Thus `sqrt(4)` is always 2 while `x^2=4` is both +2 and -2. Thus y-3 >= 0
Thus `ygt=3` .............(3)
Squaring (1) on both sides, we get
`y^2 +9 - 6y = 63 - 3y`
`y^2 - 3y - 54 = 0`
`(y-9)*(y+6) = 0`
Thus y = -6, 9.
As these both values satisfy the inequality (2) but -6 does not satisfy (3), only y=9 is acceptable.
Hence y = 9.