A voltage divider is shown in the attachment I have added, along with a formula for calculating Vout. The formula is valid and will allow you to calculate Vout for any voltage divider situation, but I tend to view the problem slightly differently. Here's how I do it:
What you've basically got here is two resistors connected in series across a supply rail. If we assume the top supply rail is 5V, bottom rail is 0V and for example let's say R1=15K and R2=10K. The situation we have then is that we have two resistors, a 15K and a 10K, connected in series across a 5V supply rail. In effect, we have 25K of total resistance connected across a 5V supply rail.
Using ohm's law, we can quite easily calculate the current that will flow under these conditions. I=V/R. I= 5/25000, I=0.2mA.
Now that we know 0.2mA flows through each of these resistors, it's quite easy to use ohm's law again to calculate Vout. With the current setup, Vout will be equal to the voltage drop across R2 and, as you probably know, V=IxR. Using that, Vout=0.2mAx10000 = +2V. The remaining 3V is of course dropped across the larger 15K resistor.
You can use the voltage divider formula provided in the attachment, or this analysis procedeure to calculate Vout for any voltage divider condition which uses just two resistors. If you're talking about a voltage divider circuit which uses more than two resistors, the divider formula as it stands will not work. However, circuit analysis will still work and that's why I tend to be in the habit of tackling these kinds of problems in this manner.
Brian