Here is a simplified approach that hopefully give you the concepts.
Assume that the base emitter voltage of the transistor is 0.7V and the gain is large enough that the base current is negligible.
First of all calculate the voltage at the junction of the 10k and 6.8k resistors as if the transistor was not present.
Vbase = 20*6.8/(10+6.8) = 8.1 volts.
AS Photon points out the transistor is in emitter follower configuration so the emitter will follow the base but be lower by the base to emitter voltage.
In that case the voltage at the emitter (Vout) will be 8.1 - 0.7 = 7.4V. This will be the output voltage Vout.
Now lets how much the base current will affect things if we have a more normal transistor.
The current through the 1K resistor is Vout/1K = 7.4/1 = 7.4mA. If we assume the Hfe of the transistor is 100 (This isa reasonable gain although many transistors these days are better than that). The base current will be 7.4/100 = 74uA.
How much will that affect the voltage at the base?
The effective impedance at the base is equals to the Thevenin equivalent resistance which is 6.8K in parallel with 10k.
This is (R1*R2)/(R1 + R2) (10*6.8)/(10+6.8) = 4.04k.
The voltage drop due to the base current will be Ib * 4.04k = 0.074*4.04 = 299mV.
So the output with this correction will be about 300mV less than our original assumption when using a very high gain transistor. i.e. 7.1V.
This approach is not 100% accurate but will be very close and is more straightforward than the full analytical method, especially when some of the parameters such as the gain of the transistor are not known and will vary significantly from unit to unit.