Request for help to solve an equation with LambertW: $ (x^2-4\,x+6) e^x =y$ I want to solve the following equation: 
$$ (x^2-4\,x+6) e^x =y \tag{1} $$
It looks a bit like the following equation: 
$$ x e^x =y \tag{2} $$
Since the solution of equation (2) is: x=LambertW(y), I think the solution of equation (1) should also use the function LambertW.
I will try to better explain what I want. I’m going to study the following function: For all x>0;
$$f(x)=(x^2-4\,x+6)\,e^x $$
$$ f’(x)=(x^2-2\,x+2)\,e^x $$
For all $x>0; x^2-2\,x+2 ≥ 1$ and $e^x$ ≥ 1 
Therefore, $f’(x) ≥1 >0 $. The function f is strictly increasing on the interval $ ]0; +∞[ $. 
Furthermore, the function f is continuous. 
Therefore, for all x>0, there is a unique y>6 such that f(x)=y. 
I know the value of y and I know how to solve the equation f(x)=y numerically. For example:
$ y=100 000; x=7.905419368254814 y=100 000 000; x=15.506081342140432$.
Does anybody know how to find the function g such that $g(y)=x $ (g is the inverse function of f, i.e. $g=f^{-1}$ ). This would provide a general formula for y in terms of x without having to solve the equation numerically. Best, Jacob Safra.

No, this is quite different.  As far as I can tell, it can't be expressed in the form $z \exp(z) = f(y)$. 
LambertW is unlikely to help here.
