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. **Set variables:** Let $a = \sqrt{x + 24}$ and $b = \sqrt{x}$. Then the equation becomes: $$a + b = 12.$$ 4. **Express $a^2$ and $b^2$:** From definitions, $$a^2 = x + 24,$$ $$b^2 = x.$$ 5. **Find $a^2 - b^2$:** $$a^2 - b^2 = (x + 24) - x = 24.$$ 6. **Use difference of squares:** $$a^2 - b^2 = (a - b)(a + b) = 24.$$ 7. **Substitute $a + b = 12$:** $$(a - b) \times 12 = 24 \implies a - b = \frac{24}{12} = 2.$$ 8. **Solve the system:** $$\begin{cases} a + b = 12 \\ a - b = 2 \end{cases}$$ Add the two equations: $$2a = 14 \implies a = 7.$$ Subtract the second from the first: $$2b = 10 \implies b = 5.$$ 9. **Recall $b = \sqrt{x}$:** $$\sqrt{x} = 5 \implies x = 5^2 = 25.$$ 10. **Check solution:** $$\sqrt{25 + 24} + \sqrt{25} = \sqrt{49} + 5 = 7 + 5 = 12,$$ which satisfies the original equation. **Final answer:** $$\boxed{25}.$$