voltage regulator problem
If you want to solve this on paper, you can use load line analysis.


Calculate the diode parameter \$I_s\$ from the known operating point of 10 mA at 0.7 V.
Plot the characteristic I-V curve of the combination of the two diodes \$I = I_s\left(\exp(\frac{V_{out}}{2 V_t})-1\right)\$. (The 2 in the denominator comes from having two diodes in series)
First find the Thevenin equivalent of the 5 V source and two resistors (2.27 V with 81.8 ohms).
Plot the inverted I-V curve for the Thevenin source circuit (\$V_{out} = 2.27\  \mathrm{V} - I(81.8\ \mathrm{\Omega}\$).


The intersection of the two curves you just drew gives the operating point Vout.

However, if you build the actual circuit, you'll find that variation between components, thermal effects, etc., will result in a slightly different operating point. Likely so much so that you would have been just as accurate to use the back-of-the-envelope result that Vout is given by two diode drops at 1.4 V.