How to solve $e^{x-2} = x$

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A friend asked me what are the solutions to this equation. I know for sure it has 2 solutions (thanks to Desmos) - (0.159, 0.159) and (3.146, 3.146).

I have no idea even where to start solving this problem.

My knowledge in mathematics is sadly high school tier so please keep it as simple as possible.

Thanks in advance!

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We need to use the function $W(x)$ which is defined to be the inverse of $xe^x$ on $(-1/e, \infty)$. This is to say, if $xe^x = y$ where $x>-1$, then $W(y)=x$. Note that for $-e^{-1}<x<0$ we have two choices for what $W(x)$ can be. So, if $x < -1$ and $xe^x=y$, we define $W_{-1}$ such that $W_{-1}(y)=x$. We have $$e^{x-2}=x$$ $$\frac 1 x e^x = e^2$$ Letting $u = -x$ gives us $$-e^2=\frac{1}{u}e^{-u} = \frac{1}{ue^u}$$ $$ue^u = -e^{-2}$$ $$\implies u = W(-e^{-2}), W_{-1}(-e^{-2})$$ $$\implies x = -W(-e^{-2}), -W_{-1}(-e^{-2})$$