1. **State the problem:** Solve the equation $$12 \times \frac{120}{n + 5} = 12 \times \frac{120}{n} - 24$$ for $n$. 2. **Write the equation clearly:** $$12 \cdot \frac{120}{n + 5} = 12 \cdot \frac{120}{n} - 24$$ 3. **Simplify the terms:** $$\frac{1440}{n + 5} = \frac{1440}{n} - 24$$ 4. **Isolate terms and clear denominators:** Multiply both sides by $n(n+5)$ to eliminate fractions: $$n(n+5) \times \frac{1440}{n + 5} = n(n+5) \times \left(\frac{1440}{n} - 24\right)$$ 5. **Cancel common factors:** $$\cancel{n}(n+5) \times \frac{1440}{\cancel{n} + 5} = n \cancel{(n+5)} \times \frac{1440}{n} - 24 n (n+5)$$ which simplifies to $$1440 n = 1440 (n+5) - 24 n (n+5)$$ 6. **Expand the right side:** $$1440 n = 1440 n + 7200 - 24 n^2 - 120 n$$ 7. **Bring all terms to one side:** $$1440 n - 1440 n - 7200 + 24 n^2 + 120 n = 0$$ which simplifies to $$24 n^2 + 120 n - 7200 = 0$$ 8. **Divide entire equation by 24 to simplify:** $$\frac{24 n^2}{24} + \frac{120 n}{24} - \frac{7200}{24} = 0$$ $$n^2 + 5 n - 300 = 0$$ 9. **Solve quadratic equation:** Use the quadratic formula: $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=5$, $c=-300$. 10. **Calculate discriminant:** $$\Delta = 5^2 - 4 \times 1 \times (-300) = 25 + 1200 = 1225$$ 11. **Find square root of discriminant:** $$\sqrt{1225} = 35$$ 12. **Calculate roots:** $$n = \frac{-5 \pm 35}{2}$$ 13. **Find each solution:** - $$n = \frac{-5 + 35}{2} = \frac{30}{2} = 15$$ - $$n = \frac{-5 - 35}{2} = \frac{-40}{2} = -20$$ 14. **Check for restrictions:** Denominators $n$ and $n+5$ cannot be zero, so $n \neq 0$ and $n \neq -5$. Both solutions $15$ and $-20$ are valid. **Final answer:** $$n = 15 \text{ or } n = -20$$