Wheatstone bridge calculation

Here is a 5 second answer to start. First, put a decoupling cap (0.1uF or something like that) between LM324 pins 4 and 11 (its just good practice). Just for testing connect LM324-1 to LM324-2 (unity gain), because an OpAmp is meant to be used with feedback.

Of course you know that you will need a differential amp sensing both sides of the bridge to get any advantage out of using the bridge.

Edit

Thanks for adding the schematic.

The output differential stage, as drawn is positive feedback. So, you will need to exchange pins 9 and 10 of the LM324 connection to have negative feedback.

Also, I didn't think about this much earlier, but the 4th channel would be best to be connected as a voltage follower with the (+) input connected to return. I know that each channel is supposed to be independent, but they are not quite really. The die will have parasitic structures that can affect all the channels. For example, parasitic SCRs are common in many monolithic ICs. So, it is always best to have all channels connected.

The whole point is to match 2 different resistor values to each other, correct?

In that case, get 2 high quality fairly closely matched resistors R1, R3.

Then irregardless of the actual values of R2 and Rx, as long as they are closely matched the voltage divider circuits R1/R2 and R3/Rx will produce a near-zero voltage Vg. The smaller the voltage, the closer the match (assuming very good R1/R3 matching). You can now use R2/Rx in your circuits and they will be pretty well matched to each other.

There is another issue with resistor matching, though: temperature coefficients. These are often specified as max values and there's no telling if two given resistors will drift at the same rate or even in the same direction.