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??
 A: 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.
A: A possibility to solve this transcendental equation could be employing Lagrange inverting series.
I have already proposed this solution in:
Lagrange Bürmann Inversion Series Example
Your equation can be rewritten:
$$x^2 +(bc-2a)x+a^2-b = be^{-cx}$$ 
replacing $cx=-z$:
$$z^2 +(bc^2+2ac)z+(a^2-b)c^2 = c^2be^{z}$$ 
Now you can factor the first term so:
$$(z-z_1)(z-z_2) = c^2be^{z}$$ 
where:
$$z_1=\frac{-(bc^2+2ac)+\sqrt{(bc^2+2ac)^2-4(a^2-b)c^2}}{2}$$
$$z_2=\frac{-(bc^2+2ac)-\sqrt{(bc^2+2ac)^2-4(a^2-b)c^2}}{2}$$
then you can write :
$$z=z_1 + \frac{c^2be^{z}}{z-z_2}$$ 
Such equation can be solved by means of Lagrange inverting series as:
$$z = z_1+ \sum_{n=1}\frac{(c^2b)^n}{n!}\left[\left(\frac{d}{dx}\right)^{n-1}\frac{e^{nx}}{(x-z_2)^n} \right]_{x=z_1}$$
developing calculations, you can find:
$$z = z_1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{nc^2be^{z_1}}{z_1-z_2}\right)^n B_{n-1}\left( \frac{-2}{n(z_1-z_2)}\right)$$
where $B_n(x)$ are the Bessel polynomials, defined as:
$$B_n(x)=\sum_{k=0}^n\frac{(n+k)!}{(n-k)!k!}\left(\frac{x}{2}\right)^k$$
See: http://en.wikipedia.org/wiki/Bessel_polynomials.
another solution can be obtained exchanging $z_1$ with $z_2$:
$$z = z_2+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{nc^2be^{z_2}}{z_2-z_1}\right)^n B_{n-1}\left( \frac{2}{n(z_1-z_2)}\right)$$.
You can recover $x$ replacing $z=-cx$.
==A numerical example==
$$x^2-1=e^{x-1}$$
gives as solutions:
$$x=1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{n}{2}\right)^n B_{n-1}\left( \frac{-1}{n}\right) = 1.78947...$$
$$x=-1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{-ne^{-2}}{2}\right)^n B_{n-1}\left( \frac{1}{n}\right) = -1.0617135...$$
I don't know if above series are linked with Lambert W function or not (my hope is that such link exists...).
For details, see (!warning! It is a preprint): 
"Generalization of Lambert W-function, Bessel polynomials and transcendental equations"
http://arxiv.org/abs/1501.00138
