Step 1 :Let the quadratic be \(x^2 + ax + b\). Then the roots are \(a + 1\) and \(b + 1\). By Vieta's formulas,
Step 2 :\begin{align*} (a + 1) + (b + 1) &= -a, \\ (a + 1)(b + 1) &= b. \end{align*}
Step 3 :From the first equation, \(a + 1 = -\frac{b}{2}\). Substituting into the second equation, we get
Step 4 :\[-\frac{b}{2} (b + 1) = b.\]
Step 5 :Since \(b\) is non-zero, we can divide both sides by \(b\), to get \(-\frac{1}{2} (b + 1) = 1\). This leads to \(b = -3\).
Step 6 :Then \(a = \frac{1}{2}\), so the roots are \(\boxed{-2,\frac{3}{2}}\).