1. **State the problem:** Solve the equation $$\frac{2}{n^2 - n - 20} = \frac{1}{n^2 - n - 20} + \frac{1}{n + 4}$$ and find the solution set. 2. **Identify the domain restrictions:** The denominators cannot be zero. - For $$n^2 - n - 20 = 0$$, factor: $$n^2 - n - 20 = (n - 5)(n + 4)$$ So, $$n \neq 5$$ and $$n \neq -4$$. - Also, $$n + 4 \neq 0 \Rightarrow n \neq -4$$. 3. **Rewrite the equation:** $$\frac{2}{(n - 5)(n + 4)} = \frac{1}{(n - 5)(n + 4)} + \frac{1}{n + 4}$$ 4. **Subtract $$\frac{1}{(n - 5)(n + 4)}$$ from both sides:** $$\frac{2}{(n - 5)(n + 4)} - \frac{1}{(n - 5)(n + 4)} = \frac{1}{n + 4}$$ 5. **Simplify the left side:** $$\frac{2 - 1}{(n - 5)(n + 4)} = \frac{1}{n + 4}$$ $$\Rightarrow \frac{1}{(n - 5)(n + 4)} = \frac{1}{n + 4}$$ 6. **Cross-multiply:** $$1 \cdot (n + 4) = 1 \cdot (n - 5)(n + 4)$$ $$n + 4 = (n - 5)(n + 4)$$ 7. **Expand the right side:** $$n + 4 = n^2 + 4n - 5n - 20$$ $$n + 4 = n^2 - n - 20$$ 8. **Bring all terms to one side:** $$0 = n^2 - n - 20 - n - 4$$ $$0 = n^2 - 2n - 24$$ 9. **Solve the quadratic:** $$n^2 - 2n - 24 = 0$$ 10. **Factor or use quadratic formula:** $$n = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-24)}}{2} = \frac{2 \pm \sqrt{4 + 96}}{2} = \frac{2 \pm \sqrt{100}}{2}$$ $$n = \frac{2 \pm 10}{2}$$ 11. **Calculate roots:** - $$n = \frac{2 + 10}{2} = \frac{12}{2} = 6$$ - $$n = \frac{2 - 10}{2} = \frac{-8}{2} = -4$$ 12. **Check domain restrictions:** - $$n = -4$$ is excluded because it makes denominator zero. - $$n = 6$$ is valid. **Final answer:** $$\boxed{\{6\}}$$