\(2x^3+7x^2+7x+2=0\\ 2\left(x^3+1\right)+7x\left(x+1\right)=0\\ 2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\\ \left(x+1\right)\left(2x^2-2x+2+7x\right)=0\\ \left(x+1\right)\left(2x^2+5x+2\right)=0\\ \left(x+1\right)\left(2x+1\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-1}{2}\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;\frac{-1}{2};-2\right\}\)