1. **State the problem:** Given the function $f(x) = \frac{x(x+1)(2x+1)}{6}$, find the value of $x$ such that $f(x) = 384$. 2. **Write the equation:** $$\frac{x(x+1)(2x+1)}{6} = 384$$ 3. **Multiply both sides by 6 to clear the denominator:** $$x(x+1)(2x+1) = 384 \times 6$$ $$x(x+1)(2x+1) = 2304$$ 4. **Expand the left side:** First, expand $x(x+1)$: $$x(x+1) = x^2 + x$$ Now multiply by $(2x+1)$: $$(x^2 + x)(2x + 1) = x^2(2x+1) + x(2x+1) = 2x^3 + x^2 + 2x^2 + x = 2x^3 + 3x^2 + x$$ 5. **Set the cubic equation:** $$2x^3 + 3x^2 + x = 2304$$ 6. **Rewrite as:** $$2x^3 + 3x^2 + x - 2304 = 0$$ 7. **Solve for integer roots by trial:** Try $x=12$: $$2(12)^3 + 3(12)^2 + 12 = 2(1728) + 3(144) + 12 = 3456 + 432 + 12 = 3900$$ (too high) Try $x=8$: $$2(512) + 3(64) + 8 = 1024 + 192 + 8 = 1224$$ (too low) Try $x=10$: $$2(1000) + 3(100) + 10 = 2000 + 300 + 10 = 2310$$ (close) Try $x=9$: $$2(729) + 3(81) + 9 = 1458 + 243 + 9 = 1710$$ (too low) Try $x=11$: $$2(1331) + 3(121) + 11 = 2662 + 363 + 11 = 3036$$ (too high) 8. **Use the exact equation or approximate root:** Since $x=10$ gives 2310 which is close to 2304, check if $x=10$ is the solution. 9. **Check original function at $x=10$:** $$f(10) = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385$$ 10. **Check $x=9$:** $$f(9) = \frac{9 \times 10 \times 19}{6} = \frac{1710}{6} = 285$$ 11. **Check $x=12$:** $$f(12) = \frac{12 \times 13 \times 25}{6} = \frac{3900}{6} = 650$$ 12. **Since $f(x)$ is increasing and $f(10) = 385$ which is close to 384, the exact solution is slightly less than 10.** 13. **Conclusion:** The value of $x$ that satisfies $f(x) = 384$ is approximately $x \approx 9.9$ (between 9 and 10). **Final answer:** $x \approx 9.9$