DC offset in closed-loop ideal op-amp
Alright, I understood it now. It's just an inverter, and you make the inputs equal (because it's ideal). So:
$$R_{2} \cdot V^{-} = R_{1}\cdot(V_{out} - V^{-})$$
where because of the ideal op-amp abstraction:
$$V^{-} = V_{off} = V^{+}$$
So you substitute this on the first equation and get
$$V_{out} = V_{out}\cdot(\frac{R_{2}}{R_{1}} + 1)$$
That was easy. But the book should explicitly say that this configuration can be dealt with as the usual inverter.