Consider the following equation,
$$\sqrt{\frac{\sqrt{x!}\times \sqrt{x!!}\times \sqrt{x!!!}}{\sqrt{\left( x-2 \right)!+x}}}=12$$ I tried first to eliminates all radicals using squring both sides to get the following , $$\frac{\sqrt{x!}\times \sqrt{x!!}\times \sqrt{x!!!}}{\sqrt{\left( x-2 \right)!+x}}={{12}^{2}}\Leftrightarrow \frac{\sqrt{x!\times x!!\times x!!!}}{\sqrt{\left( x-2 \right)!+x}}=\frac{x!\times x!!\times x!!!}{\left( x-2 \right)!+x}={{12}^{4}}$$
Now how will continue attacking this problem?
Let's say we have next to no intuition for the size of a potential solution, and don't want to try out small numbers by hand. Then we can use the Gamma function in an extension of the multifactorial to make this a continuous function so that a computer can more-easily help us hunt for the root.
By the calculations in the OP, with the equation with $12^4$, we can see at the very least that the solution probably isn't bigger than $12^2$, and likely much less. Let's just start looking around $150$: Wolfram|Alpha says that $45.1$ would be a better guess. If you start at $45$, it says that something around $6.2$ would be better. If you start at $6$, it's still thinking around $6.2$ (but these continuous functions are hard to approximate), so we should probably check around $6$ or $7$.
If we happen to choose $7$ first, it says the value is about $18.5$. And if we ask for the value at 6, we find that it is indeed $12$.