How to find point of intersection between quadratic and exponential equation How can I find the point of intersection between a function like $2^x$ and $x^2$?
I know you have to equate them but I don't know what to do after that.

As said in answers, there are two trivial roots $x=2,x=4$ and the extra one is given in terms of Lambert function.
If you do not want (or cannot) use Lambert function, only numerical methods could do  the work. Probably the simplest is Newton which, starting at a "reasonable" guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us try $f(x)=2^x-x^2$ and, because I am lazy, let me start with $x_0=0$. The iterates will be  $-1.44270$, $-0.897065$, $-0.773470$, $-0.766685$, $-0.766665$ which is the solution for six significant figures.
Starting with $x_0=-1$ would have provided as iterates : $-0.786923$, $-0.766843$, $-0.766665$.
The second sequence does not show any overshoot of the solution because $$f(-1)f''(-1)=1-\frac{\log ^2(2)}{4}>0$$ while $$f(0)f''(0)=\log ^2(2)-2<0$$ If I properly remember, this condition corresponds to some theorem by Darboux.
Edit
You can generate a good starting point from Taylor series built at $x=0$ $$f(x)=1+x \log (2)+x^2 \left(\frac{\log ^2(2)}{2}-1\right)+O\left(x^3\right)$$ If you just keep the first order, you then have $x_0=-\frac{1}{\log (2)}$ we already obtained in the first path; but, using the second order and solving the quadratic, you obtain $$x_0=\frac{\sqrt{4-\log ^2(2)}-\log (2)}{\log ^2(2)-2} \approx -0.778455$$
