How can I determine the equation for the output voltage?
One way to simplify the problem is to use the Thevenin equivalent circuits for those voltage dividers you have there. That is, you could draw the circuit in this manner:







simulate this circuit &ndash; Schematic created using CircuitLab

Here $$V_{th1}=\frac{R_1}{R_{sensor}+R_1}V_{cc}$$ 
\$R_1\$ is just your 680 ohm resistor in series with your 500 ohm potentiometer. \$V_{cc}\$ is the supply voltage.

Your \$R_{th1}\$ is simply:
$$R_{th1}=R_1||R_{sensor} $$

The same procedure is done for \$V_{th2}\$ and \$R_{th2}\$. 

You should get \$ V_{th2}=\frac{V_{cc}}{2}\$ and \$R_{th2}=\frac{R}{2}\$. Where \$R=1.2k\Omega\$.

Now, you can use \$V^-=V^+\$ and start your opamp analysis. In the end, you should obtain the equation for the opamp as a differential amplifier. I'll let you do the math, but your final solution for \$V_{out}\$ should be (if I didn't mess up somewhere):

$$ V_{out}=V_{th2}-\frac{R_f}{R_{th1}}(V_{th1}-V_{th2})$$

Where \$R_f\$ is your \$5k\Omega\$ feedback potentiometer.