A difference amplifier with a non-grounded non-inverting input
The simple and fairly universal answer is to just use Ohms Law and solve the resulting equations for each node voltage.  Of course with experience one can often just look at a simple circuit like this and write down the final equation.

Let V+ be the voltage on the non-inverting input.
Let V- be the voltage on the inverting input of the op-amp.
Let Vin1 be one of the input voltages.
Let Vin2 be one of the input voltages.
Let Vout be the output voltage.
Let R1 be a resistor connected between V- and Vin1
Let R2 be a resistor connected between V+ and Vin2
Let Rf be a resistor connected between Vout and V-
Let I1 be the current through R1.
Let I2 be the current through R2.
Let If be the current through Rf.

Lets start by analyzing the non-inverting input.
If we define I2 as being positive when current flows from Vin2 to V+ then...

I2 = (Vin2 - V+) / R2 , by ohms law

Solving for V+ gives...

V+ = Vin2 - I2 * R2.

Real op-amps have some input bias current Ib+ flowing into their non-inverting input.  Note that I2 = Ib+ (since there is no where else for the current to go), therefore...

V+ = Vin2 - Ib+ * R2.

For precision op-amps the input bias current is typically nano-amps or pico-amps.  So for R2=100 ohms, the difference between V+ and Vin2 would be in the micro-volt range.

For high speed op-amps Ib+ may be in the micro-amp range, so the difference between V+ and Vin2 would be in the low millivolt range.  In either case the difference is typically ignored as an error signal that factors into the overall tolerance of the design.  

The non-inverting input also has some input bias current Ib-.  An additional error voltage is developed at V- due to Ib- and the impedance of the attached resistors.    

An op-amp has an output that is proportional to the difference between V+ and V-.  Therefore if the error at V+ is exactly equal to the error at V- then the errors will cancel.  Many datasheet often specify that the input offset current (the difference between Ib+ and Ib-) is much smaller than either Ib+ or Ib-.  If the resistance seen by V+ and V- are equal then similar errors will be produced and cancel each other thus increasing the accuracy of the output.

For an ideal op-amp Ib+ is assumed to be 0, and in that case Vin2 = V+ regardless of weather or not R2 is there.

Now lets analyze the inverting input.

For a real op-amp Vout = Avol * (Vin+ - Vin-), where Avol is the open loop gain of the op-amp.  

I1 = (Vin1 - V+) / R1, ignoring bias current.

For an ideal op-amp G is assumed infinite, therefore Vin+ = Vin- so...

I1 = (Vin1 - Vin2) / R1

Note that...

If = (V- - Vout) / Rf

Note that If = I1 therefore...

(Vin1 - Vin2) / R1 = (V- - Vout) / Rf

Also since V- = V+ = Vin2 we have...

(Vin1 - Vin2) / R1 = (Vin2 - Vout) / Rf

Solving for Vout gives...

Vout = Vin2 + (Vin2 - Vin1) * Rf/R1