8. If the function $f(x)$ defined on $\mathbf{R}$ satisfies $f^{\prime}(x)-2 f(x)-4>0, f(0)=-1$, then the solution to the inequality $f(x)>e^{2 x}-2$ is
Let $g(x)=e^{-2 x}(f(x)+2)$, then $g(0)=1$. From $g^{\prime}(x)=e^{-2 x}\left[f^{\prime}(x)-2 f(x)-4\right]>0$ we know that $g(x)$ is monotonically increasing in $(-\infty,+\infty)$, thus $g(x)>1 \Leftrightarrow x>0$, hence $f(x)>e^{2 x}-2 \Leftrightarrow x>0$.