I know the answer is (0,2) U (2, infinity)
f(x) = squareroot(4x-3)
(x^2)-4
Do you set the denominator to zero? I am not 100% sure.
$\displaystyle \displaystyle f(x) = \frac{\sqrt{4x-3}}{x^2-4}$
Are you sure it's not $\displaystyle \displaystyle \left[\frac{3}{4},2\right)\cup (2,\infty)$?
You are on the right track; set the denominator to zero because when it is zero, f is undefined (so exclude any such x from the domain). Also, solve $\displaystyle 4x-3<0$ because that will make $\displaystyle \sqrt{4x-3}$ non-real.