jmc

The function $|x|$ is difficult to deal with directly. Instead we work by cases: $x\ge 0$ and $x<0$.

If $x\ge 0$ then $|x|=x$, and we can find the difference by subtracting $$x-(-x^2-3x-2)=x^2+4x+2=(x+2{)}^2-2.$$This function is always increasing as $x$ varies over the nonnegative numbers, so this is minimized at $x=0$. The minimum value on $x\ge 0$ is $$(0+2{)}^2-2=2.$$If $x<0$ then $|x|=-x$ and we can find the difference by subtracting: $$(-x)-(-x^2-3x-2)=x^2+2x+2=(x+1{)}^2+1.$$This quadratic is minimized at $x=-1$, and the minimum value is $$(-1+1{)}^2+1=1.$$Since the minimum value on negative numbers is less than the minimum value on the nonnegative numbers, the minimum value for the difference is $\boxed{1}$.