(x2-3)½-(x-3)½=0,
(x2-3)½=(x-3)½,
x2-3=x-3,
x2-x=0,
x(x-1)=0, so x=0 or 1.
However, neither side of the original equation can be evaluated in real terms, because each would need the square root of a negative number. In real terms, then, there is no solution.
If complex numbers are permitted then the solutions are valid:
When x=0: (-3)½=(-3)½, so i√3=i√3, which is true.
When x=1: (-2)½=(-2)½, so i√2=i√2, which is true.