The solution found is wrong , because , On either it's sides it is surrounded by same voltage , so no current flows through 8ohms and 2ohms , so no voltage drops , which means Vo is just 4V ( if you use voltage divider in the positive terminal circuit ) but again if V1 is 4V that would mean some current is flowing inside op amp , since none is coming from the Vo side , so this circuit could be wrong .
It's just a simplification to make ideal calculations with. When there's a feedback resistor in the circuit, both inputs will have very nearly the same voltage. It's the slight difference in the voltages that acts as the signal to be amplified. That difference is so slight, though, that it can be ignored in calculations (a real op-amp has a gain of 10000 times or more, so a 100mV input voltage difference might give a 1000V output). An ideal op-amp has infinite gain, so the voltage difference is infinitely small.
