The number of real solutions of the equation e^x=2x^2 is ____.

Roots of $e^x=2x^2$ are not at $x=0$ (left side is $1$, right side $0$), so divide by $x^2$ and study $h(x)=\dfrac{e^x}{x^2}=2$. Its derivative is $h'(x)=\dfrac{e^x(x-2)}{x^3}$, which is positive for $x<0$ (numerator and $x^3$ both negative), so on $(-\infty,0)$ the function $h$ increases from $0$ to $+\infty$ as $x$ goes from $-\infty$ to $0^-$, crossing $2$ exactly once. For $x>0$, $h'<0$ on $(0,2)$ and $h'>0$ on $(2,\infty)$, with minimum $h(2)=\dfrac{e^2}{4}\approx1.85<2$ and $h\to\infty$ at both ends of $(0,\infty)$, so $h=2$ has exactly two positive roots. The total is $1+2=3$.