1. **State the problem:** Solve for $x$ in the equation $$\sqrt{\frac{(x+2)!}{x!}} = \sqrt{3!} \times 7.$$\n\n2. **Recall factorial and square root properties:** The factorial $n!$ is the product of all positive integers up to $n$. Also, $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.\n\n3. **Simplify the left side:** Using the factorial property, $$\frac{(x+2)!}{x!} = (x+2)(x+1)$$ because $$\frac{(x+2)!}{x!} = \frac{(x+2)(x+1)x!}{x!} = (x+2)(x+1).$$\n\n4. **Rewrite the equation:** $$\sqrt{(x+2)(x+1)} = \sqrt{3!} \times 7.$$\n\n5. **Calculate $3!$:** $$3! = 3 \times 2 \times 1 = 6.$$\n\n6. **Substitute $3!$ and combine square roots:** $$\sqrt{(x+2)(x+1)} = \sqrt{6} \times 7 = 7\sqrt{6}.$$\n\n7. **Square both sides to eliminate the square root:** $$\left(\sqrt{(x+2)(x+1)}\right)^2 = (7\sqrt{6})^2,$$\nwhich simplifies to $$ (x+2)(x+1) = 49 \times 6 = 294.$$\n\n8. **Expand the left side:** $$x^2 + x + 2x + 2 = x^2 + 3x + 2.$$\n\n9. **Set up the quadratic equation:** $$x^2 + 3x + 2 = 294.$$\n\n10. **Bring all terms to one side:** $$x^2 + 3x + 2 - 294 = 0,$$\nwhich simplifies to $$x^2 + 3x - 292 = 0.$$\n\n11. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$\nwhere $a=1$, $b=3$, and $c=-292$.\n\n12. **Calculate the discriminant:** $$\Delta = 3^2 - 4 \times 1 \times (-292) = 9 + 1168 = 1177.$$\n\n13. **Find the roots:** $$x = \frac{-3 \pm \sqrt{1177}}{2}.$$\n\n14. **Approximate $\sqrt{1177}$:** $$\sqrt{1177} \approx 34.31.$$\n\n15. **Calculate the two possible values for $x$:**\n$$x_1 = \frac{-3 + 34.31}{2} = \frac{31.31}{2} = 15.655,$$\n$$x_2 = \frac{-3 - 34.31}{2} = \frac{-37.31}{2} = -18.655.$$\n\n16. **Check for valid factorial inputs:** Since factorials are defined for non-negative integers and the original expression involves factorials of $x$ and $x+2$, $x$ must be a non-negative integer.\n\n17. **Test integer values near $15.655$:** The closest integer is $16$.\n\n18. **Verify $x=16$:**\n$$\frac{(16+2)!}{16!} = \frac{18!}{16!} = 17 \times 18 = 306,$$\n$$\sqrt{306} \approx 17.4929,$$\nwhile $$\sqrt{3!} \times 7 = \sqrt{6} \times 7 \approx 2.4495 \times 7 = 17.1465.$$\n\n19. **Verify $x=15$:**\n$$\frac{(15+2)!}{15!} = \frac{17!}{15!} = 16 \times 17 = 272,$$\n$$\sqrt{272} \approx 16.4924,$$\nwhich is less than $17.1465$.\n\n20. **Verify $x=14$:**\n$$\frac{(14+2)!}{14!} = \frac{16!}{14!} = 15 \times 16 = 240,$$\n$$\sqrt{240} \approx 15.4919,$$\nwhich is less than $17.1465$.\n\n21. **Since $x=15.655$ is not an integer, and factorials require integer inputs, the exact solution is not an integer.**\n\n22. **Conclusion:** The solution to the equation is $$x = \frac{-3 \pm \sqrt{1177}}{2}.$$\nConsidering the domain of factorials, the approximate valid solution is $$x \approx 15.655.$$