Number of Real solutions of $e^{x^2}=ex$ 
Find a number of real  solutions of the following equation.
   $$e^{x^2}=ex$$

Need a pure calculus approach..
 A: $$(e^{x^2})''=2(2x^2+1)e^{x^2}>0,$$
which says that $f(x)=e^{x^2}$ is a convex function.
Thus, our equation has at most two real roots.
But $1$ is a root and it's obvious that there is a root on $(0,1)$,
which says that the answer is $2$.
A: If you take the natural log of your equation you can arrange it to read
$$x^2-1 = \ln x.$$
The two sides are easy to graph and intersect at $x=1$.  Noting that the left side is always concave up and the right is always concave down shows there is a second solution (around 0.45.)
A: Maybe help you .  
take $$f(x)=e^{x^2}-ex$$so $$f'(x)=2xe^{x^2}-e=0\\x=0.761\\$$

one of roots is $x=1 $ but smaller root is between $$(0,0.761)$$ .
You can find it numerically .
A: 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)
$$a_{20}=0.450763653$$
A: I believe the following is a short proof sketch that the answer is 2.  Clearly an integer solution is $x=1$.  You can argue based on the geometry of the curves $f(x)=e^{x^{2}}$ and $g(x)=ex$ that the intersection between the two contains either zero, one, or two points.  
Well you already know it is either one or two, since $x=1$ is a solution.  Therefore, where the calculus comes in, is showing that $ex$ is not the tangent line to $f(x)=e^{x^{2}}$ at $x=1$.  There then must be two real roots.
EDIT: Michael Rozenberg's argument above about convexity should be used to replace my handwaving about the 'shape of the functions.'  
