Circuit analysis with an LED and a voltage divider
Voltage dividers aren't especially useful for LEDs.

LEDs are current driven devices, not voltage driven devices.  Voltage dividers provide a voltage.
Voltage dividers aren't good when there's a load attached to them.  The load changes the resistance and therefore the divider ratio.

To make a somewhat stable voltage divider, you need at least about 10 times the current flowing through the resistors as through the load.  If your load needs 20 milliamperes, then you will need at least 200 milliamperes through both resistors.  At 12V, that makes the total of the two resistors have to be less than \$ \frac {12V}{.2A} = 60 \Omega\$.
If you use that for your 20 milliampere rated LED, though, you will destroy the LED.  You'd only have that relatively small top resistor between the LED and the 12V - the LED would get more than 200 milliamperes and promptly burn out.

Your LED didn't &quot;pop&quot; because the current is limited by R1.  The maximum current would be \$\frac {12V}{R1 + R2} = 0.022 A\$  That's at or about the rated value of a typical LED.  Part of the current goes through R2 and part of the current goes through the LED.  R2 diverts enough current from the LED to keep it from burning out.
In your circuit, the voltage at the junction of R1 and R2 will always be the forward voltage of the LED.



simulate this circuit &ndash; Schematic created using CircuitLab
The voltage across the LED depends almost entirely on R1, which control the current to R2 and the LED.  R2 reduces the current the LED a little.
If you build the circuit, you will never get the calculated voltage (from the voltage divider equations) across the LED.  You will get a voltage that depends on the current through the LED, and the current to voltage curve of the LED (which you can find in the LED datasheet.)

A simple series resistor is all it takes to power an LED from a voltage source.



simulate this circuit
The datasheet of the LED will tell you the rated forward current (\$I_f\$) and the approximate forward voltage (\$V_f\$) for the rated forward current.
What you do is calculate a resistor value that will limit the current to the rated \$I_f\$ given your supply voltage (\$V_S\$.)
The resistor value is given by: \$R = \frac{V_S-V_f}{I_f}\$
A typical blue LED has a forward voltage of about 3V at 20 milliamperes.  You want to operate it from 12V.
\$ R = \frac{12V-3V}{0.02A} = 450 ohms\$