jmc

By the quadratic formula, $x=\frac{-b+\sqrt{b^2-4c}}{2},\frac{-b-\sqrt{b^2-4c}}{2}$. The difference of these is $\frac{2\sqrt{b^2-4c}}{2}=\sqrt{b^2-4c}$. Setting this equal to $|b-2c|$, it follows that (after squaring) $b^2-4c=(b-2c{)}^2=b^2+4c^2-4bc$. Thus $$0=4c^2+4c-4bc=4c(c-b+1).$$As $c\ne 0$, it follows that $c=\boxed{b-1}$.