1. **State the problem:** Solve the equation $$45 = 3x + \sqrt{x - 5}$$ for $x$. 2. **Isolate the square root term:** $$\sqrt{x - 5} = 45 - 3x$$ 3. **Important rule:** The expression under the square root must be non-negative, so: $$x - 5 \geq 0 \implies x \geq 5$$ Also, since the square root is equal to $45 - 3x$, the right side must be non-negative: $$45 - 3x \geq 0 \implies x \leq 15$$ 4. **Square both sides to eliminate the square root:** $$\left(\sqrt{x - 5}\right)^2 = (45 - 3x)^2$$ $$x - 5 = (45 - 3x)^2$$ 5. **Expand the right side:** $$(45 - 3x)^2 = 45^2 - 2 \cdot 45 \cdot 3x + (3x)^2 = 2025 - 270x + 9x^2$$ 6. **Rewrite the equation:** $$x - 5 = 2025 - 270x + 9x^2$$ 7. **Bring all terms to one side:** $$0 = 2025 - 270x + 9x^2 - x + 5$$ $$0 = 9x^2 - 271x + 2030$$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{271 \pm \sqrt{271^2 - 4 \cdot 9 \cdot 2030}}{2 \cdot 9}$$ Calculate the discriminant: $$271^2 = 73441$$ $$4 \cdot 9 \cdot 2030 = 73080$$ $$\sqrt{73441 - 73080} = \sqrt{361} = 19$$ 9. **Find the two possible solutions:** $$x = \frac{271 \pm 19}{18}$$ 10. **Calculate each:** $$x_1 = \frac{271 + 19}{18} = \frac{290}{18} = \frac{145}{9} \approx 16.11$$ $$x_2 = \frac{271 - 19}{18} = \frac{252}{18} = 14$$ 11. **Check domain restrictions:** Recall $x \geq 5$ and $x \leq 15$ from step 3. $x_1 \approx 16.11$ is not in the domain, so discard. $x_2 = 14$ is in the domain, so keep. 12. **Verify solution $x=14$ in original equation:** $$3 \cdot 14 + \sqrt{14 - 5} = 42 + \sqrt{9} = 42 + 3 = 45$$ Correct. **Final answer:** $$\boxed{14}$$