1. **Problem:** Solve the equation $$\frac{x^2 - 9}{x + 3} = x^2 - 15$$. 2. **Formula and rules:** When dealing with rational expressions, first check for values that make the denominator zero (excluded values). Here, $x + 3 \neq 0 \Rightarrow x \neq -3$. 3. **Step 1:** Factor the numerator: $$x^2 - 9 = (x - 3)(x + 3)$$ 4. **Step 2:** Simplify the left side: $$\frac{(x - 3)(x + 3)}{x + 3} = x - 3 \quad \text{for} \quad x \neq -3$$ 5. **Step 3:** Substitute back into the equation: $$x - 3 = x^2 - 15$$ 6. **Step 4:** Rearrange to form a quadratic equation: $$0 = x^2 - x - 12$$ 7. **Step 5:** Factor the quadratic: $$x^2 - x - 12 = (x - 4)(x + 3)$$ 8. **Step 6:** Solve for $x$: $$x - 4 = 0 \Rightarrow x = 4$$ $$x + 3 = 0 \Rightarrow x = -3$$ 9. **Step 7:** Check excluded values. Since $x = -3$ makes the denominator zero, exclude it. 10. **Final answer:** $$\boxed{x = 4}$$ This matches the provided solution.