How to find number of solutions to $2^x=x^2$ without Graphing Find number of real solutions to $2^x=x^2$ without plotting graph:
I considered $f(x)=2^x -x^2$ 
$$f'(x)=2^x \ln 2-2x=0$$
we get again a transcendental equation. Any good approach please

$a_{0}2^x+b_{0}$ may have at most one zero. Before that it is negative, after that zero, if it exists, it is positive.
Based on just this remark, $a_{1}2^x+b_{1}x$ may have at most two zeros, since its derivative $a_{0}2^x+b_{0}$ is forcing it to behave similar to a parabola, if the derivative has one zero. Again, based on the essential property of $a_{1}2^x+b_{1}x$, $a_{2}2^x+b_{2}x^2$ may have at most three zeros (since it may have at most two extreme values).
That this is really the case is easily seen from its behavior around $2$.
$2^2-2^2=0$, and it is has a negative slope at $2$ since at $1$, $2^1-1^2>0$ and at $3$, $2^3-3^2<0$
Now
$\lim\limits_{x \to -\infty}2^x-x^2<0$ and $\lim\limits_{x \to \infty}2^x-x^2>0$ and since $2^x-x^2$ is continuous there has to be one zero before and one zero after $2$.
