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 substituting $4!$:** $$\sqrt{x} + 24 + \sqrt{x} = 12.$$\n\n4. **Combine like terms:** $$2\sqrt{x} + 24 = 12.$$\n\n5. **Isolate the square root term:** $$2\sqrt{x} = 12 - 24 = -12.$$\n\n6. **Divide both sides by 2:** $$\cancel{2}\sqrt{x} = \frac{-12}{\cancel{2}} \Rightarrow \sqrt{x} = -6.$$\n\n7. **Analyze the result:** The square root of a real number $x$ cannot be negative, so $\sqrt{x} = -6$ has no real solution.\n\n8. **Conclusion:** There is no real value of $x$ satisfying the equation. If complex numbers are considered, then $\sqrt{x} = -6$ implies $x = (-6)^2 = 36$.\n\n**Final answer:** No real solution. Complex solution: $x = 36$.