12. Given that $x$ is a real number, and satisfies $5^{2 x+1}+3125=5^{5 x-x^{2}}$, then the sum of the minimum and maximum values of $x$ is $\qquad$.
Solve:
$$
5^{2 x+1}+5^{5}=5^{5 x-x^{2}} \Rightarrow 5^{x^{2}-3 x+1}+5^{x^{2}-5 x+5}=1 .
$$

Since $(4-x)^{2}-3(4-x)+1=x^{2}-8 x+16+3 x-12+1=x^{2}-5 x+5$, let $f(x)=5^{x^{2}-3 x+1}+5^{x^{2}-5 x+5}$, then $f(4-x)=f(x)$. Therefore, the sum of the minimum and maximum values of $x$ that satisfy the equation is 4.