\(3x^2+10x-8=0\\ \Leftrightarrow3x^2+12x-2x-8=0\\ \Leftrightarrow3x\left(x+4\right)-2\left(x+4\right)=0\\ \Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-4;\dfrac{2}{3}\right\}\)