It is known how to solve equations of the form $e^{-cx}=a_0(x-r)$ in terms of the Lambert $W$-function. Namely, according to Wikipedia, the solution is given by $$ x=r+\frac{1}{c}W\bigg(\frac{ce^{-cr}}{a_0}\bigg). $$
Is there a similar way to solve equations of the form $$ e^{-cx^2}=a_0(x-r), $$ where say $a_0,c,r>0$? In this case, there is indeed a unique solution with positive $x$ value, for the left hand side is positive and the right hand side is linear and increasing in $x$ and is negative at $x=0$. Without the $r$, one could reduce to the above solution by squaring. I'd be interested in if there are either special functions that are reasonably well-understood to solve this, or if there are any reasonable approximations on the value of $x$ in terms of $c,a_0,$ and $r$. Thanks!