Problem with quadratics and exponentiation I was bored and started solving the following equation:
$$2^x = x^2$$
I can see two solutions:
$x = 2$ and $x = 4$.
WolframAlpha tells me there is one more, but I can hardly get the two I mentioned above (at which I arrived with empirical methods).
I fear this might be outside the realm of my Maths skills (up to multivariable calculus).
 A: If you are asking (I assume you are) about how to get these two results analytically, I am almost certain that there is no way to get a closed solution to the equation $a^x = x^a$, except for special cases like $a=2$ in your case. In that case, using plots to have a look at how many solutions you can expect is useful:
http://www.wolframalpha.com/input/?i=plot+2%5Ex-x%5E2
You can see that there is another solution for some negative value of $x$, which you probably cannot get without numeric methods.
For the positive solutions, some things can actually be said without resorting to numeric methods.
You can show that for $x<2$, the function $2^x - x^2$ is positive, since it is increasing at $x=0$, decreasing at $x=2$ and is concave on $[0,2]$. 
For $x>2$, you can show that because the function is convex, it can have at most two zeroes on $$[2,\infty).
Together, these two points show that the only positive solutions to your equation are $2$ and $4$
A: In Wolfram, use 
solve x^2 - 2^x = 0
To find the negative solution to be about -0.766665
https://www.wolframalpha.com/input/?i=solve+x%5E2+-+2%5Ex+%3D+0
