Domain of Composite Rational Functions

Let $y = \frac{2x - 3}{2x^2 - 4x - 6}$. $y(2x^2 - 4x - 6) = 2x - 3 \implies 2y x^2 - 2(2y + 1)x - (6y - 3) = 0$. For $x \in \mathbb{R}$, discriminant $D \ge 0$: $D = 4(2y + 1)^2 - 4(2y)(-(6y - 3)) \ge 0$ $4(4y^2 + 4y + 1) + 8y(6y - 3) \ge 0$ $16y^2 + 16y + 4 + 48y^2 - 24y \ge 0$ $64y^2 - 8y + 4 \ge 0$. Wait, let's re-calculate $D$ carefully: $D/4 = (2y+1)^2 - 2y(-6y+3) = 4y^2 + 4y + 1 + 12y^2 - 6y = 16y^2 - 2y + 1 > 0$ for all $y$. Wait! The domain excludes points where denominator is 0: $2x^2 - 4x - 6 = 0 \implies x^2 - 2x - 3 = 0 \implies x = 3, -1$. Since $x \neq 3, -1$, check values of $y$ excluded or bounded.