jmc

Dividing the equation by $x^2,$ we get $$2x^2+x-6+\frac{1}{x}+\frac{2}{x^2}=0.$$Let $y=x+\frac{1}{x}.$ Then $$y^2=x^2+2+\frac{1}{x^2},$$so $x^2+\frac{1}{x^2}=y^2-2.$ Thus, we can re-write the equation above as $$2(y^2-2)+y-6=0.$$This simplifies to $2y^2+y-10=0.$ The roots are $y=2$ and $y=-\frac{5}{2}.$

The roots of $$x+\frac{1}{x}=2$$are 1 and 1. The roots of $$x+\frac{1}{x}=-\frac{5}{2}$$are $-2$ and $-\frac{1}{2}.$

Thus, the roots of $2x^4+x^3-6x^2+x+2=0$ are $\boxed{1,1,-2,-\frac{1}{2}}.$