Voltage subtractor using op-amp
By inspection (using superposition), I get that (assuming ideal op-amp)

$$V_{OUT} = 0.75\cdot \left(1 + \frac{100}{150||300}\right) - 9\cdot\frac{100}{300}$$

$$= 0.75\cdot 2 - 9\cdot\frac{1}{3} = -1.5 \mathrm{V}$$

But, in your schematic, the op-amp cannot produce a negative output voltage. Assuming that it can drive the output to zero, we have

$$V_- = 9\cdot \frac{150||100}{150||100 + 300} = 1.5 \mathrm{V}$$

as you observe.