I gave this equation $2^x=1+x^2$ with the $1$st zero is $x_1=0$ and the $2$nd zero is $x_2=1$. (easy reading)
Now I want to calculate more zeros using Rolle's theorem, and I've rearranged the function for this: $$f(x)=1+x^2-2^x$$
and formed the first two derivatives: $f'(x)=2x-\ln(2)*2^x$ and $f''(x)=2-\ln(2)^2*2^x$
So and from here it fails now. I know what Rolle's theorem says. Between two zeros of the function there is a zero of the derivative, and if $f$ is twice differentiable, then between three zeros of the function there are two zeros of the first derivative and one zero of the second derivative.
Also, I saw that $f(4) > 0$ and $f(5) < 0$. Therefore, in the interval (4,5) there is at least one more zero of $f$ according to the intermediate value theorem.
But how do I determine furthermore precise real solutions of the equation?
Note that most often Rolle's theorem isn't used to prove the existence of solutions, but to show that not too many solutions exist (that said, one can in some cases use Rolle's theorem to prove the existence of solutions to $f(x)=0$ if one knows that $f(x)$ is the derivative of some function with appropriate zeros).
You found $3$ solutions using the intermediate value theorem, so we can prove these are the only ones:
The second derivative $f''(x)=2-(\ln 2)^2\cdot 2^x$ has only one root because $f''(x)=0$ implies $2^x = \frac{2}{(\ln 2)^2}$ and $y=2^x$ is injective. Therefore by Rolle's theorem $f'(x)=0$ has at most $2$ solutions (if it had more, $f''(x)=0$ would have more than one solution), and therefore $f(x)=0$ has at most $3$ solutions (if it had more, $f'(x)=0$ would have more than $2$ solutions).