jmc

Substituting $f^{-1}(x)$ into our expression for $f$ we get $$f(f^{-1}(x))=2f^{-1}(x)-5.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $$x=2f^{-1}(x)-5.$$or $$f^{-1}(x)=\frac{x+5}{2}.$$We want to solve the equation $f(x)=f^{-1}(x)$, so $$2x-5=\frac{x+5}{2}.$$or $$4x-10=x+5.$$Solving for $x$, we find $x=\boxed{5}$.