TITLE: Number of Real solutions of $e^{x^2}=ex$
QUESTION [6 upvotes]: Find a number of real  solutions of the following equation.
   $$e^{x^2}=ex$$

Need a pure calculus approach..

REPLY [2 votes]: Disclaimer: Related but does not answer the question.
(It does however, tell you how to find the roots in terms of the Lambert W function)
Start by squaring both sides to get
$$e^{2x^2}=e^2x^2$$
Divide both sides by $-e^{2x^2-2}/2$ to get
$$-2e^{-2}=-2x^2e^{-2x^2}$$
Apply the Lambert W function to both sides to get
$$-2x^2=W_k(-2e^{-2})$$
Or,
$$x=\pm\sqrt{-\frac12W_k(-2e^{-2})}$$
Since $-e^{-1}<-2e^{-2}<0$, there are two real valued branches of the Lambert W function, so we have 4 possible solutions:
$$x=\pm\sqrt{-\frac12W_{k}(-2e^{-2})},\quad k=-1,0$$
Trivially, the solution can't be negative, since $e^{x^2}>0$, so this reduces down to
$$x=\sqrt{-\frac12W_{k}(-2e^{-2})},\quad k=-1,0$$
For $k=-1$, we find that
$$W_{-1}(-2e^{-2})=-2$$
And for $k=0$, we find that
$$W_0(-2e^{-2})=-\sum_{n=1}^\infty\frac{n^{n-1}}{n!}(2e^{-2})^n\approx-0.4064$$
Which gives respective solutions
$$x=1,0.4508$$
One may we wish to exploit different representations of our solution, such as
$$\sqrt{-\frac12W_0(-2e^{-2})}=\exp\left(-\frac12W_0(-2e^{-2})-2\right)=\frac1{2e^2}\sum_{n=0}^\infty\frac{(n+0.5)^{n-1}}{n!}(2e^{-2})^n$$
Or, perhaps if you prefer numerical iteration methods,
$$a_0=0.5,~~a_{n+1}=e^{a_n^2-1}$$
Upon which I obtain (with my unperfect calculator)
$$a_{20}=0.450763653$$