Solution in terms of Lambert $W$ function Is it possible to solve equation of the following form using Lambert $W$ function.
$$(x-a)^2 = b(e^{-cx} - cx + 1).$$ 
If not, can it be solved using any other special function??

I do not think so because of the square in the lhs.
If it had been instead $$(x-a)=b(e^{-c x}-c x+1)$$ the solution would have been $$x= \frac{W\left(\frac{b c e^{-\frac{c (a+b)}{b c+1}}}{b c+1}\right)}{c}+\frac{a+b}{b
   c+1}$$ Basically, if the equation can be written (after any series of transformations) $$A+By+C\log(D+Ey)=0$$ then the solution for $y$ would have been expressed using Lambert function
$$y=\frac{C}{B} W\left(\frac{B e^{\frac{B D-A E}{C E}}}{C E}\right)-\frac{D}{E}$$
For the solution of $$(x-a)^2=b(e^{-c x}-c x+1)$$ I think that only numerical methods will give you results.
