1. **State the problem:** Solve for $x$ in the equation $$\sqrt{\frac{(x+2)!}{x!}} = \sqrt{3! \times 7}.$$\n\n2. **Recall factorial properties:** The factorial of a number $n$ is $n! = n \times (n-1) \times \cdots \times 1$. Also, $$\frac{(x+2)!}{x!} = (x+2)(x+1)$$ because the $x!$ terms cancel out all factors up to $x!$.\n\n3. **Rewrite the equation using this property:**\n$$\sqrt{(x+2)(x+1)} = \sqrt{3! \times 7}.$$\n\n4. **Calculate $3!$:**\n$$3! = 3 \times 2 \times 1 = 6.$$\n\n5. **Substitute $3!$ back:**\n$$\sqrt{(x+2)(x+1)} = \sqrt{6 \times 7} = \sqrt{42}.$$\n\n6. **Square both sides to eliminate the square root:**\n$$\left(\sqrt{(x+2)(x+1)}\right)^2 = (\sqrt{42})^2$$\n$$\Rightarrow (x+2)(x+1) = 42.$$\n\n7. **Expand the left side:**\n$$x^2 + x + 2x + 2 = 42$$\n$$x^2 + 3x + 2 = 42.$$\n\n8. **Bring all terms to one side:**\n$$x^2 + 3x + 2 - 42 = 0$$\n$$x^2 + 3x - 40 = 0.$$\n\n9. **Factor the quadratic:**\n$$x^2 + 3x - 40 = (x + 8)(x - 5) = 0.$$\n\n10. **Solve for $x$:**\n$$x + 8 = 0 \Rightarrow x = -8$$\n$$x - 5 = 0 \Rightarrow x = 5.$$\n\n11. **Check for valid factorial inputs:** Factorials are defined for non-negative integers, so $x = -8$ is invalid.\n\n12. **Final answer:**\n$$\boxed{5}.$$