1. **State the problem:** Solve for $x$ in the equation $$\sqrt{x} + 4! + \sqrt{x} = 12.$$\n\n2. **Recall the factorial value:** $4! = 4 \times 3 \times 2 \times 1 = 24$.\n\n3. **Rewrite the equation:** Substitute $4!$ with 24:\n$$\sqrt{x} + 24 + \sqrt{x} = 12.$$\n\n4. **Combine like terms:** There are two $\sqrt{x}$ terms, so\n$$2\sqrt{x} + 24 = 12.$$\n\n5. **Isolate the square root term:** Subtract 24 from both sides:\n$$2\sqrt{x} + \cancel{24} - \cancel{24} = 12 - 24,$$\nwhich simplifies to\n$$2\sqrt{x} = -12.$$\n\n6. **Divide both sides by 2:**\n$$\cancel{2}\sqrt{x} = \frac{-12}{\cancel{2}},$$\nso\n$$\sqrt{x} = -6.$$\n\n7. **Analyze the result:** The square root of a real number $x$ cannot be negative. Since $\sqrt{x} = -6$ is impossible for real $x$, there is no real solution.\n\n**Final answer:** There is no real value of $x$ satisfying the equation.