A theoretical approach would be:
Since the ideal op-amp input impedance is infinite, you can pretend the negative input (Vn) does not affect the rest of the circuit. So, the lower part of the circuit becomes a voltage divider and, since both resistors have the same value, the middle voltage becomes the average of both ends.
Vn = (Vo + Vref) / 2
By that same argument, you can also say that the positive input (Vp) is the same as Vdac (since no current flows through the resistor).
Vp = Vdac
Now, if you take the basic op-amp gain equation and isolate Vn, you get:
Vo = A * (Vdac - Vn)
Vo = A*Vdac - A*Vn
A*Vn = A*Vdac - Vo
Vn = Vdac - Vo/A
Then you can join both equations and work your way into isolating the output (Vo):
(Vo + Vref) / 2 = Vdac - Vo/A
Vo + Vref = 2*Vdac - 2*Vo/A
Vo + 2*Vo/A = 2*Vdac - Vref
(A*Vo + 2*Vo) / A = 2*Vdac - Vref
Vo * (A + 2) = (2*Vdac - Vref) * A
Vo = (2*Vdac - Vref) * A / (A + 2)
That last equation seems to be a little complicated, but since the ideal op-amp gain is really large (infinite), you may consider that
A ~= A + 2 (for A >>>> 0), then it becomes very simple:
Vo = (2*Vdac - Vref) * A / A
Vo = 2*Vdac - Vref
That's it, you've got Vo as a function of Vdac and Vref and I believe it makes it easier to see what Vo does.