# How to solve $x^2=x!$

If $$x^2=x!$$ , what are the values of $$x$$? And this equation too, if $$x!=120$$, we know that here $$x=5$$ because $$5!=5×4×3×2×1=120$$. This seems something like back calculation. Is there any process to solve this type of equation normally?

$$x! >x(x-1)(x-2) >x^{2}$$ for $$x \geq 4$$ so you only have to check if the equation holds for $$x=1,2,3$$.