Noting that the coefficients are symmetric. Because it is obvious that $x\ne 0$, we can divide both sides by $x^2$ and rearrange these terms: $$2\Big(x^2 +\frac{1}{x^2}\Big)+3\Big(x+\frac{1}{x}\Big)-16=0$$

Let $y=x+\frac{1}{x}$, the above equation is equivalent to:$$2(y^2-2)+3y-16=0\implies y_{1,2}=\frac{5}{2}, -4$$

When $y=\frac{5}{2}$, $x+\frac{1}{x}=\frac{5}{2}\implies x_{1,2}=\frac{1}{2}, 2$. When $y=-4$, $x+\frac{1}{x}=-4\implies x_{3,4}=-2\pm\sqrt{3}$. Here there are four roots: $\boxed{\frac{1}{2}, 2, -2+\sqrt{3}, -2-\sqrt{3}}$.