Extending Output of a fixed Linear Voltage Regulator
The output voltage \$V_O\$ will be whatever the regulator voltage is plus whatever the voltage is at the top of \$R2\$.

\$V_O = V_{XX} + (I_{R2}) * R2\$

So, to determine the voltage at the top of \$R2\$ you need to know the current going through it, \$I_{R2}\$.

That will be \$I_Q\$, the quiescent current of the device plus whatever current goes through \$R1\$. 

\$V_O = V_{XX} + (I_Q + I_{R1}) * R2\$

Since \$R1\$ has the regulator voltage across it, we know:

\$I_{R1} = V_{XX}/R1\$

So, substituting for \$I_{R1}\$, we get:

=> \$V_O = V_{XX} + (I_Q + V_{XX}/R1) * R2\$

=> \$V_O =  V_{XX} + V_{XX}.R2/R1 + I_Q.R2\$

=> \$V_O= V_{XX}.(1 + R2/R1) + I_Q.R2\$



Note: This circuit should never be used if the input voltage exceeds the maximum input voltage, \$V_I\$, of the regulator chosen. I have seen folks try to use this method in an attempt to "shoe-horn" the device to work at higher voltages, and, although once in a steady state it will work ok, sooner or later the initial start-up, or shut-down voltages will kill the device if you do not surround it with protection devices.