\(16\left(2-3x\right)+x^2\left(3x-2\right)=0\\ \left(-16\right)\left(3x-2\right)+x^2\left(3x-2\right)=0\\ \left(3x-2\right)\left(x^2-16\right)=0\\ \left(3x-2\right)\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{2}{3};4;-4\right\}\)