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, $$\frac{(x+2)!}{x!} = (x+2)(x+1)$$ because $$\frac{(x+2)!}{x!} = \frac{(x+2)(x+1)x!}{x!} = (x+2)(x+1).$$\n\n3. **Rewrite the equation using this simplification:** $$\sqrt{(x+2)(x+1)} = \sqrt{3!} \times 7.$$\n\n4. **Calculate $3!$:** $$3! = 3 \times 2 \times 1 = 6.$$\n\n5. **Substitute $3!$ value:** $$\sqrt{(x+2)(x+1)} = \sqrt{6} \times 7 = 7\sqrt{6}.$$\n\n6. **Square both sides to eliminate the square root:** $$\left(\sqrt{(x+2)(x+1)}\right)^2 = \left(7\sqrt{6}\right)^2.$$\n\n7. **Simplify both sides:** $$ (x+2)(x+1) = 7^2 \times 6 = 49 \times 6 = 294.$$\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:** $$x^2 + 3x + 2 - 294 = 0 \Rightarrow x^2 + 3x - 292 = 0.$$\n\n11. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$ where $a=1$, $b=3$, $c=-292$.\n\n12. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 1 \times (-292) = 9 + 1168 = 1177.$$\n\n13. **Calculate the roots:** $$x = \frac{-3 \pm \sqrt{1177}}{2}.$$\n\n14. **Approximate $\sqrt{1177}$:** $$\sqrt{1177} \approx 34.31.$$\n\n15. **Find 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 domain restrictions:** 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. **Check 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$$\sqrt{306} \approx 17.4929,$$\n$$\sqrt{3!} \times 7 = \sqrt{6} \times 7 \approx 2.449 \times 7 = 17.143.$$\n\n19. **Check $x=15$:**\n$$\frac{(15+2)!}{15!} = \frac{17!}{15!} = 16 \times 17 = 272,$$\n$$\sqrt{272} \approx 16.492,$$\n$$\sqrt{6} \times 7 \approx 17.143.$$\n\n20. **Check $x=14$:**\n$$\frac{(14+2)!}{14!} = \frac{16!}{14!} = 15 \times 16 = 240,$$\n$$\sqrt{240} \approx 15.491,$$\n$$\sqrt{6} \times 7 \approx 17.143.$$\n\n21. **Check $x=13$:**\n$$\frac{(13+2)!}{13!} = \frac{15!}{13!} = 14 \times 15 = 210,$$\n$$\sqrt{210} \approx 14.491,$$\n$$\sqrt{6} \times 7 \approx 17.143.$$\n\n22. **Check $x=17$:**\n$$\frac{(17+2)!}{17!} = \frac{19!}{17!} = 18 \times 19 = 342,$$\n$$\sqrt{342} \approx 18.493,$$\n$$\sqrt{6} \times 7 \approx 17.143.$$\n\n23. **Since the exact value is $\sqrt{294} \approx 17.146$, the closest integer $x$ that satisfies the equation is $x=15$ or $x=16$ but neither equals exactly.**\n\n24. **Conclusion:** The exact solutions are $$x = \frac{-3 \pm \sqrt{1177}}{2}.$$\n\n**Final answer:** $$\boxed{x = \frac{-3 \pm \sqrt{1177}}{2}}.$$