Did you consider doing what you weretoldto do? That is, "use your calculator", as opposed to "solving analytically". Thiscannotbe solved "analytically", using elementary functions, which is what you are trying to do. (The answer is NOT "2 and some decimal". There is, in fact, two simple solutions you could get by 'inspection', one that's a bit more complicated. But, ignoring that...)

Instead, consider graphing 2^x and x^2 on your (graphing) calculator and "zooming" in on the point at which they cross.

Or do this. Let f(x)= 2^x- x^2 which will be equal to 0 if x satisfies 2^x= x^2. If x= -1, f(-1)= 2^-1- (-1)^2= 1/2- 1= -1/2. Since f(1) is positive, f(-1) is negative, and f is continuous, there must be a solution somewhere between -1 and 1. Where? Don't know! But since it is easy to divide by 2, lets look halfway between: (-1+ 1)/2= 0/2= 0. If x= 0, f(0)= 2^0- 0^2= 1. That is positive so there must be a solution somewhere between -1 and 0. Again, try halfway between (just because it is easy). (-1+ 0)/2= -1/2. (2^{-/2}- (-1/2)^2= [tex]1/\sqrt{2}- 1/4= 0.4571, approximately. That is still positive (but close to 0) so there must be a solution between -1 and -1/2. Again, half way between is -3/4= -.75. f(-3/4)= 2^{-3/4}- (-3/4)^2= 0.5946- 0.5625= 0.0321. Again, that is positive so now we know there is a solution between -.75 and -1. Continue until you have enough accuracy.

(There are two positive solutions but those are integers so I would not use this method!)

The simplest way to solve this equation is what I first suggested: graph and and zoom in on the intersections.