1. Stating the problem: Solve for $x$ in the equation $$\sqrt{x} + 4! + \sqrt{x} = 12$$. 2. Recall that $4!$ (4 factorial) means $4 \times 3 \times 2 \times 1 = 24$. 3. Substitute $4!$ with 24 in the equation: $$\sqrt{x} + 24 + \sqrt{x} = 12$$ 4. Combine like terms $\sqrt{x} + \sqrt{x} = 2\sqrt{x}$: $$2\sqrt{x} + 24 = 12$$ 5. Isolate the term with $\sqrt{x}$ by subtracting 24 from both sides: $$2\sqrt{x} + \cancel{24} - \cancel{24} = 12 - 24$$ $$2\sqrt{x} = -12$$ 6. Divide both sides by 2 to solve for $\sqrt{x}$: $$\cancel{2}\sqrt{x} / \cancel{2} = -12 / 2$$ $$\sqrt{x} = -6$$ 7. Since $\sqrt{x}$ represents the principal (non-negative) square root, it cannot be negative. Therefore, there is no real solution for $x$. Final answer: \textbf{No real solution} because $\sqrt{x}$ cannot be negative.