\(3x-4x^2+6-8x>x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2>0\Leftrightarrow\left(5x-1\right)\left(x+2\right)>0\)
TH1 : \(\left\{{}\begin{matrix}5x-1>0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x>-2\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{5}\)
TH2 : \(\left\{{}\begin{matrix}5x-1< 0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x< -2\end{matrix}\right.\Leftrightarrow x< -2\)