Let the quadratic be \[x^2 + ax + b.\]Then the roots are $ a + 1 $ and $ b + 1 $. By Vieta's formulas, \begin{align*} (a + 1) + (b + 1) &= -a, \\ (a + 1)(b + 1) &= b.\end{align*}From the first equation, $ a + 1 = -\frac{b}{2} $. Substituting into the second equation, we get \[-\frac{b}{2} (b + 1) = b.\]Since $ b $ is non-zero, we can divide both sides by $ b, $ to get $ -\frac{1}{2} (b + 1) = 1 $. This leads to $ b = -3 $. Then $ a = \frac{1}{2}, $ so the roots are $ \boxed{-2,\frac{3}{2}} $.