How to solve $x^2 = e^x$ The question is to find $x$ in:

\begin{equation*}
x^2=e^x
\end{equation*}



I know Newton's method and hence could find the approx as $x\approx -0.7034674225$ from 

\begin{equation*}
x_{n+1} = x_n - \dfrac{x_n^2-e^{x_n}}{2x_n-e^{x_n}}
\end{equation*}

According to WolframAlpha:


They also say that $x=-2W(\dfrac{1}{2})$ which shows that it can be solved using some Lambert-W function...Can anyone tell me how to do this?

Thanks a lot!
P.S. - I studied a li'l bit of Lambert-W ... So i guess a detailed explanation would not be needed ... just the initial steps! 
 A: The Lambert W function is the inverse of $xe^x$. 
We want to find the inverse of $e^x x^{-2}$, dividing by $x^2$. 
Now:
\begin{align}
y &= x^{-2} e^x \\
y^{-0.5} &= x e^{-0.5x} &&\vee y^{-0.5} = -x e^{-0.5x} \\
-0.5y^{-0.5} &= -0.5x e^{-0.5x} &&\vee 0.5y^{-0.5} = -0.5x e^{-0.5x} \\
W(-0.5y^{-0.5}) &= -0.5x  &&\vee W(0.5y^{-0.5}) = -0.5x  \\
x &= -2W(-0.5y^{-0.5}) &&\vee x = -2W(0.5y^{-0.5}) \\ \\
\end{align}
Now we have y=1, so $x=-2W(-0.5)$ or $x=-2W(0.5)$.
The first one is complex, so only the second one remains as real solution. 
A: $$
x^2=e^x\implies x/2=\pm\tfrac12e^{x/2}\implies-x/2\,e^{-x/2}=\pm\tfrac12
$$
Therefore,
$$
x=-2\mathrm{W}\!\left(\pm\tfrac12\right)
$$
Since $\mathrm{W}(x)$ is real only for $x\ge-\frac1e$, we only have one real solution:
$$
x=-2\mathrm{W}\!\left(\tfrac12\right)=-0.70346742249839165205
$$
A: Write the equation as $x^2e^{-x} = 1$. Then $x^2e^{-x} = 4\left(\left(-\frac{x}{2}\right)e^{-\frac{x}{2}}\right)^2$.
A: You can write $\mathrm{f(x)=e^{-x}.x^2}$
Now differentiate this function w.r.t x.
$$\mathrm{f`(x)=e^{-x}(2x-x^2)}$$.
Now set $$\mathrm{f'(x)=0}$$.
It yields the solutions $$\mathrm{x=0 and x=2}$$.
The values of f(x) at these x values are 0 and $\mathrm{4/e^2}$ respectively. Obviously $\mathrm{4/e^2}$ is less than 1.
Now plot the graph of this function. Find the number of intersection points of this function and $\mathrm{y=1}$.
