How to do DC Analysis for a Common Source Biasing Circuit using NMOS
The nature of these MOSFET DC analysis problems is that you end up with a quadratic equation and you have to pick the solution which satisfies the assumption you made at the beginning. 

You assumed saturation and I will do the same. Let's see if there is a solution which satisfies our assumption.

First, for saturation, in the end, we have to verify that two conditions are met:
 $$V_{GS} &gt; V_t  $$ $$ V_{DS}\geq V_{GS}-V_t$$

So, for saturation, as you have it in your post: $$ I_D=k(V_{GS}-V_t)^2$$I am absorbing the \$ \frac{1}{2}\$ into \$ k\$.

What I will do now is replace \$V_{GS}\$ with some equation that includes \$I_D\$.
 $$V_{GS} = V_G-V_S=V_G-I_DR_2 $$ 
If we now plug this into our \$I_D\$ equation, we have $$I_D=k(V_G-I_DR_2-V_t)^2 $$ The math gets a little messy, I hope I didn't miss anything. But if you expand the quadratic term and collect the \$I_D\$ terms, you should end up with something like this: $$ I_{D}^2(kR_2^2)+I_{D}(2kR_2V_t-2kV_GR_2-1)+(kV_G^2-2kV_GVt+kV_t^2)=0$$

I used MATLAB to solve for \$I_D\$ and I get the following roots:
$$I_{D1}=3mA $$
$$I_{D2}=2.9mA $$

For the first solution, \$V_{GS}=1.5V&lt;V_t\$. You can rule out this solution since it won't satisfy our assumption of saturation. It would mean the MOSFET is cut off.

The second solution gives \$V_{GS}=1.6V&gt;V_t\$. That's a good start, but we still need to satisfy the second condition of $$ V_{DS}\geq V_{GS}-V_t$$

$$V_{DS}=V_{DD}-I_D(R_1+R_2)=0.3V $$
$$ V_{GS}-V_t =0.05V$$
And \$0.3V\geq 0.05V\$. So the second condition is met and the MOSFET is in saturation. 

I am not sure if I used the right units for \$k\$. Usually this is expressed in terms of \$\frac{mA}{V^2}\$ or \$\frac{\mu A}{V^2}\$. So I used \$\frac{A}{V^2}\$ since no units were given. This could change the region of operation for the MOSFET (could be in triode, for example).

All in all, that is the approach that I would take. I hope it helps!