1. We are asked to solve the inequality $$\frac{x^2 + 3}{2x - 8} > 0$$. 2. Important points to consider are where the numerator and denominator are zero because these points divide the number line into intervals. 3. The numerator is $$x^2 + 3$$, which is always positive since $$x^2 \geq 0$$ and adding 3 makes it strictly positive for all real $$x$$. 4. The denominator is $$2x - 8$$. Set it equal to zero to find critical points: $$2x - 8 = 0$$ $$2x = 8$$ $$x = 4$$ 5. The denominator changes sign at $$x=4$$. For $$x < 4$$, $$2x - 8 < 0$$, and for $$x > 4$$, $$2x - 8 > 0$$. 6. Since the numerator is always positive, the sign of the fraction depends on the denominator: - For $$x < 4$$, denominator is negative, so fraction is negative. - For $$x > 4$$, denominator is positive, so fraction is positive. 7. The inequality $$\frac{x^2 + 3}{2x - 8} > 0$$ holds when $$x > 4$$. 8. Also, $$x=4$$ is excluded because the denominator is zero there. 9. Final solution: $$\boxed{x > 4}$$