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 function $n!$ is the product of all positive integers up to $n$. Also, $$\frac{(x+2)!}{x!} = (x+1)(x+2)$$ because $$\frac{(x+2)!}{x!} = \frac{(x+2)(x+1)x!}{x!} = (x+1)(x+2).$$\n\n3. **Rewrite the equation using this simplification:** $$\sqrt{(x+1)(x+2)} = \sqrt{3! \times 7}.$$\n\n4. **Calculate $3!$:** $$3! = 3 \times 2 \times 1 = 6.$$\n\n5. **Substitute $3!$ back:** $$\sqrt{(x+1)(x+2)} = \sqrt{6 \times 7} = \sqrt{42}.$$\n\n6. **Square both sides to eliminate the square root:** $$\left(\sqrt{(x+1)(x+2)}\right)^2 = (\sqrt{42})^2$$\n$$\Rightarrow (x+1)(x+2) = 42.$$\n\n7. **Expand the left side:** $$x^2 + 3x + 2 = 42.$$\n\n8. **Bring all terms to one side:** $$x^2 + 3x + 2 - 42 = 0$$\n$$x^2 + 3x - 40 = 0.$$\n\n9. **Factor the quadratic:** Find two numbers that multiply to $-40$ and add to $3$. These are $8$ and $-5$.\n$$x^2 + 8x - 5x - 40 = 0$$\n$$x(x + 8) - 5(x + 8) = 0$$\n$$(x - 5)(x + 8) = 0.$$\n\n10. **Solve for $x$:**\n$$x - 5 = 0 \Rightarrow x = 5$$\n$$x + 8 = 0 \Rightarrow x = -8.$$\n\n11. **Check for valid factorial inputs:** Factorials are defined for non-negative integers. $x = -8$ is invalid.\n\n**Final answer:** $$\boxed{5}.$$