If two of the roots of $2x^3-3x^2-3x+2=0$  are differ by 3 then roots are

$\left(x-2\right)\left(2x^2+x-1\right)$

Take $2x^2+x-1=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{-1\pm \sqrt{1-4\left(2\right)\left(-1\right)}}{4}=\frac{-1\pm 3}{4}=-1,\frac{1}{2}$

Roots are $-1,\frac{1}{2},2$