Number of Real solutions of $e^{x^2}=ex$

$$(e^{x^2})''=2(2x^2+1)e^{x^2}>0,$$ 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 $2$.

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.)

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 .