R=(Vs-Vf)/I where R is the required series resistor, Vs is the supply voltage, Vf is the total forward voltage drop, and I is the diode current. so for a pair of LEDs that operate at 3.5V forward drop (each) at 100mA, and a 12V supply... the total forward drop of two diodes in series is 7V, so subtract from 12V and you have 5V 5V/100mA=50 ohms. the power dissipation of the resistor should be 5V*100mA=0.5W. with a safety margin, and using a standard value from the E24 series of resistors gives you a 47 ohm 1 watt resistor. the current will be slightly higher, but not enough to be an issue, although if you want slightly less current use a 51ohm 1 watt resistor. you will find the 47 ohm more commonly available.