9. (16 points) Find all real solutions to the equation $2^{x}-x^{2}-28=0$.


II. 9. Notice that, $2^{6}=6^{2}+28$.
Therefore, $x=6$ is a solution to the equation.
Below, we prove that this is also the only real solution.
By $2^{x}=x^{2}+28>2^{4} \Rightarrow x>4$.
Consider the function
$$
f(x)=2^{x}-x^{2}-28(x \in[4,+\infty))
$$

monotonicity.
We can prove that $f^{\prime}(x)=2^{x} \ln 2-2 x$ is always positive on the interval $(4,+\infty)$.
In fact, $\ln 2=\frac{1}{2} \ln 4>\frac{1}{2}$.
When $x>4$,
$$
\begin{array}{l}
f^{\prime \prime}(x)=2^{x}(\ln 2)^{2}-2>2^{4}\left(\frac{1}{2}\right)^{2}-2>0 \\
\Rightarrow f^{\prime}(x) \text { is an increasing function on the interval }(4,+\infty) \\
\Rightarrow f^{\prime}(x)>f^{\prime}(4)=2^{4} \ln 2-2 \times 4>0 .
\end{array}
$$

Therefore, the function $f(x)$ is an increasing function and has at most one zero.
Hence, $x=6$ is the only real solution.