The quadratic $x^2 + \frac{3}{2} x - 1$ has the following property: the roots, which are $\frac{1}{2}$ and $-2,$ are one less than the final two coefficients.  Find a quadratic with leading term $x^2$ such that the final two coefficients are both non-zero, and the roots are one more than these coefficients.  Enter the roots of this quadratic.
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 $\boxed{-2,\frac{3}{2}}.$