There is no bias.
If driving the LCD from an MCU or logic chip giving 5V output, then measuring with respect to the chip ground:
Signal 1 goes | +5V | 0V | +5V | 0V | +5V | 0V |
Signal 2 goes | 0V | +5V | 0V | +5V | 0V | +5V |
But if the LCD is not connected to this 0V reference point, then what
is the reference point for voltage measurement? Answer - whichever connection to the LCD you care to choose. The one for signal 1 or the one for signal 2. It doesn't matter which one you choose - but for the sake of the discussion, let's pick signal 1. Put your black lead on that point and leave it there.
The result is that
this reference point will follow the voltage excursions of the output signal and will always measure 0V with respect to it (stick your red lead on that point. Yes, you will have both the red and the black leads on the same point - but that's typically what a 0V measurement entails!). It doesn't matter what the voltage is with respect to
anything else or how that changes over time - it will always be 0V.
Now move the red lead to the connection for the signal 2. What voltage does it measure now?
In DC terms, it will change over time. When signal 1 is low, signal 2 will be high - and since signal 2 is 5V higher than signal 1, the (instantaneous) voltage measurement will be +5V
Now, when signal 1 is high, signal 2 will be low - and since signal 2 is 5V lower than signal 1, the (instantaneous) voltage measurement will be -5V.
Thus, the voltage change on one pin will be either +5V or -5V
with respect to the other pin.
This is, in effect,
exactly the same as an AC square wave with +5V and -5V levels.
As others have said, this is working identically to how an H-bridge operates ... so if you are still lost, it will be worth finding a reference on that.