As you leaned in the intro to opamps, when under negative feedback, the voltage difference across its inputs will be close to 0v. This is achieved via compensation from the opamp output and the feedback loop.
In the simplest way to achieve it is a configuration known as the inverting amplifier. In this configuration, the non inverting input is tied directly to ground, and a feedback loop is made using a resistor connected between inverting input and output.
Another resistor is used to connect the signal source to the amplifier, since the inverting input will be at the same potential by action of the feedback loop, it would be connected to ground and no signal would get to the opamp to get amplified.
The voltage in through the resistor will cause a current going in the direction of the inverting input. Since one of the properties of the op amp is that its inputs draw virtually no current, or at least it will try not to draw current by pulling the output voltage towards a more negative value, in order to create a voltage across the feedback resistor that will draw the same amount of current as what's trying to go through the input resistor.
The math behind this action:
Vrin = Vin - Vinv
The inverting terminal is at the same potential as the non inverting, which is tied to ground, so:
Vrin = Vin
Then separate Vrin into current times voltage, according to ohm's law:
IinRin = Vin => Iin = Vin/Rin
Now you get an equation for the current in. Since the input will not draw current, we have that
Iin = Ifb, Ifb is the feedback current.
Ifb = (Vinv - Vout)/Rfb
Again, Vinv is tied to ground similar to the non inverting, so
Ifb = (0 - Vout)/Rfb => -Vout/Rfb
We equal both currents to get an equation in terms of only voltages and resistors
Iin = Ifb => Vin/Rin = -Vout/Rfb
The variable of interest is Vout, so rewrite it in terms of Vout
(Vin/Rin)Rfb = -Vout => -Vin(Rfb/Rin) = Vout
From this last equation you can see that the output voltage will be an inverted version of Vin multiplied by the ratio of the input resistor and the feedback resistor; increasing the input resistor gives less gain, while increasing the feedback resistor increases gain.
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