Usually I solve this kind of problems in this way.
The "positive feedback" forces the opamp to deliver either +Vcc or -Vee (i.e. the positive and negative supply voltages) to its output.
So there are two cases:
- Output voltage is +Vcc. This condition is held until the V+ terminal has a greater voltage than V-. So I calculate V+, V- and say "the threshold from high to low is when"...
- Output voltage is -Vee. This condition is held until the V+ terminal has a lower voltage than V+. I proceed exactly in the same way as above.
One numerical example.
simulate this circuit – Schematic created using CircuitLab
Let's assume the output is at +5V. It will stay at +5V until
V+ > V-, so
V+ = (Vout + V2) / 2 = V2/2 + 2.5V
V- = V1/2
The value changes, consequently, when
V- > V+, so when
V1/2 > V2/2 + 2.5V
V1 > V2 + 5V
So the threshold from high to low is
V1 = V2 + 5V
As for the other, when the output is -Vee (0V) the condition is
V+ = (Vout + V2) / 2 = V2/2
V- = V1/2
It will stay in this condition until
V+ < V-; consequently it will change status when
V+ > V-
V1 < V2
So the low-to-high threshold is
V1 = V2
So, let's assume V2 constant. When V1 raises above V2 by more than 5V the output will switch to high (Vcc). Then it has to go lower than V2 in order for the output to become low.
Of course I didn't solve your exercise, because you need it in order to understand if you understood it