Solve $t\, e^t=t+2$ for $t>0$. I need to solve the following equation:
$$t\, e^t=t+2\, ,$$
for $t>0$. Do I need to use Lambert W function, or there is some other method?
Thanks!

PS. I know that any equation of the form $A+Bt+C\log(D+Et)=0$ has solutions which, if they exist, can be expressed using Lambert function. But I do not think that $t\, e^t=t+2$ can be rewrite in this way... am I wrong?

 A: I do not believe that this can be solved using the Lambert $W$ function. When one attempts to put it in the form that you specified, one ends up with
$$\ln\bigg(\frac{t}{t+2}\bigg)+t=0$$
If we let $\frac{t}{t+2}=u$, then we have
$$\ln(u)+\frac{2u}{1-u}=0$$
$$(1-u)\ln(u)+2u=0$$
So if you can find some way of solving equations in the form
$$A+Bt+(C+Dt)\ln(E+Ft)=0$$
then this would be possible... however, I am afraid that you may have to attack this problem without the Lambert $W$.
If you would like to approximate, I would recommend using iteration.
A: As I said in a comment, there is a solution involving the generalized Lambert function. If you look here,  C. B. Corcino and R. B. Corcino  proved
that in the asymptotic expression of the  generalized Bell numbers
one needs to solve the equation $$x\,e^x + r x=n$$ which is exactly your case. The remaining of the paper addresses the more general case of $$e^{-cx}=a \frac{ \prod_{i=1}^n (x-t_i)} {\prod_{i=1}^m (x-s_i) }$$
Loooking at the posted equation, from a purely numerical point of view, Newton method will converge quite fast using $t_0=1$ observed by inspection. The successive iterates would be 
$$\left(
\begin{array}{cc}
 0 & 1.000000000 \\
 1 & 1.063499184 \\
 2 & 1.060100663 \\
 3 & 1.060090320
\end{array}
\right)$$ which is the solution for ten significant figures.
We also could use $[1,n]$ Padé approximants built around $t=1$ and get explicit approximations of the solution. For example
$$t_{(1)}=1+\frac{2 (3-e) (2 e-1)}{2+e+5 e^2}\approx 1.0599978$$
$$t_{(2)}=1+\frac{3 (3-e) \left(2+e+5 e^2\right)}{2 \left(-3+9 e+15 e^2+8 e^3\right)}\approx 1.0600921$$
$$t_{(3)}=1+\frac{8 (3-e) \left(-3+9 e+15 e^2+8 e^3\right)}{24-21 e+291 e^2+265 e^3+65 e^4}\approx 1.0600903$$
A: If you rearrange the original expression to get:
$$t=t \exp(t)-2$$
it is easy to see that the solution is a fixed point for function $f(t)=t \exp(t)-2$, that is a contraction map in the region $t\in (-\infty, 0)$. Therefore a solution can be found by the sequence $$t_{i+1}=f(t_i),\ \ i=0,1,2,...$$
that is guaranteed to converge to a fixed point if we start with $t_0$ in the interval in which $f(t)$ is a contraction map. Doing that will lead to $t^*\approx -2.238643$ after a few iterations.
A: Here is a plot:

The solutions to
$$t{e^t} = t + 2$$
are
$$t \approx -2.23865$$
and
$$t \approx 1.06009.$$
Credits: I used WolframAlpha.
