Solution of $x^2e^x = y$ The other day, I came across the problem (or something that reduced to the problem):
Solve for $x$ in terms of $y$ and $e$: $$x^2e^x=y$$
I tried for a while to solve it with logarithms, roots, and the like, but simply couldn't get $x$ onto one side by itself without having $x$ on the other side too.
So, how can I solve this, step-by-step?
More generally, how can I solve equations that involve both polynomials (e.g. $x^2$, $x^3$) and exponentials (e.g. $e^x$,$10^x$)?
EDIT - I now remember why I this question came up.  I was reading something about complexity theory (the basics: P, NP, NP-hard, etc.), and I got to a part that talked about how polynomial time is more efficient than exponential time.  So, I decided to take a very large polynomial function and a very small exponential function and see where they met.  Hence, I had to solve an equation with both polynomials and exponentials, which I figured could reduce to $x^2e^x=y$.

Solution with Lambert W:
$$
x^2 e^x=y
\\
x e^{x/2} = \sqrt{y}
\\
\frac{x}{2}\;e^{x/2} = \frac{\sqrt{y}}{2}
\\
\frac{x}{2} = W\left(\frac{\sqrt{y}}{2}\right)
\\
x = 2\;W\left(\frac{\sqrt{y}}{2}\right)
$$
One solution for each branch of the W function.  
Other solutions by taking the other square-root:
$$
x = 2\;W\left(\frac{-\sqrt{y}}{2}\right)
$$
Is there "no point" in these solutions?  Perhaps.  There are known properties of W.  Your CAS may already have W coded.
