My way is similar to that of

**earboth**:

Equating the numerator and denominator to zero, we find two critical numbers:

Dividing the real number line at these critical value gives us 3 intervals. Since neither of the critical numbers is a root of even multiplcity, we know the sign of the expression will alternate across the intervals. We pick a test number from the first (leftmost) interval:

test value

expression is negative, so our solution will be:

We use an open interval since the inequality is strict. If it were weak, we would still have to exclude 2 to avoid division by zero.