1. Stating the problem: Solve the inequality $$x^2 > 3x + 4$$. 2. Rearrange all terms to one side to compare to zero: $$x^2 - 3x - 4 > 0$$. 3. Factor the quadratic expression: $$x^2 - 3x - 4 = (x - 4)(x + 1)$$. 4. The inequality becomes: $$(x - 4)(x + 1) > 0$$. 5. This product is greater than zero when both factors are positive or both are negative. 6. Case 1: Both positive: $$x - 4 > 0 \implies x > 4$$ $$x + 1 > 0 \implies x > -1$$ Since $x > 4$ is more restrictive, solution for case 1 is: $$x > 4$$. 7. Case 2: Both negative: $$x - 4 < 0 \implies x < 4$$ $$x + 1 < 0 \implies x < -1$$ Since $x < -1$ is more restrictive, solution for case 2 is: $$x < -1$$. 8. Combine solutions from both cases: $$x < -1 \quad \text{or} \quad x > 4$$. Final answer: $$x < -1 \text{ or } x > 4$$.