1. **State the problem:** Solve for $x$ in the equation $$\sqrt{x + 4!} + \sqrt{x} = 12.$$ 2. **Recall the factorial:** $4! = 4 \times 3 \times 2 \times 1 = 24$. Substitute this into the equation: $$\sqrt{x + 24} + \sqrt{x} = 12.$$ 3. **Isolate one square root:** Let $a = \sqrt{x + 24}$ and $b = \sqrt{x}$. Then the equation is $a + b = 12$. 4. **Express $a$ in terms of $b$:** $$a = 12 - b.$$ 5. **Square both sides to eliminate the square root:** $$a^2 = (12 - b)^2.$$ 6. **Substitute back $a^2$ and $b^2$:** $$x + 24 = (12 - \sqrt{x})^2 = 144 - 24\sqrt{x} + x.$$ 7. **Simplify the equation:** $$x + 24 = 144 - 24\sqrt{x} + x.$$ 8. **Cancel $x$ from both sides:** $$\cancel{x} + 24 = 144 - 24\sqrt{x} + \cancel{x}.$$ 9. **Simplify further:** $$24 = 144 - 24\sqrt{x}.$$ 10. **Isolate the square root term:** $$24\sqrt{x} = 144 - 24.$$ 11. **Calculate the right side:** $$24\sqrt{x} = 120.$$ 12. **Divide both sides by 24:** $$\cancel{24}\sqrt{x} = \frac{120}{\cancel{24}} = 5.$$ 13. **Solve for $\sqrt{x}$:** $$\sqrt{x} = 5.$$ 14. **Square both sides to find $x$:** $$x = 5^2 = 25.$$ 15. **Check the solution:** $$\sqrt{25 + 24} + \sqrt{25} = \sqrt{49} + 5 = 7 + 5 = 12,$$ which satisfies the original equation. **Final answer:** $$x = 25.$$