Voltage biasing problem
Look over your schematic. Other than the supply rails, there are three nodes: \$V_G\$, \$V_D\$, and \$V_S\$. Also, you can construct the fact that \$V_{GS}=V_G-V_S\$. Correct?

You should be able to develop the trivial voltage divider result for the \$\ref{vdiv}\$:

$$V_G=V_{CC}\frac{R_2}{R_1+R_2}\label{vdiv}\tag{gate voltage}$$

That already solves one of your nodes. The others are similarly trivial. you have the \$\ref{vs}\$ and the \$\ref{vd}\$:

$$V_D=V_{CC}-R_D\cdot I_D\label{vd}\tag{drain voltage}$$

$$V_S=R_S\cdot I_D\label{vs}\tag{source voltage}$$

The \$\ref{vd}\$ isn't terribly important, right now. It will fall out for sanity testing later, once you figure out the value for \$I_D\$.

At this point, all you do is enter into your starting equation:

$$\begin{align*}
I_D&amp;=K\cdot\left(V_{GS}-V_T\right)^2\\\\
&amp;=K\cdot\left(V_G-V_S-V_T\right)^2\\\\
&amp;=K\cdot\left(V_{CC}\frac{R_2}{R_1+R_2}-R_S\cdot I_D-V_T\right)^2
\end{align*}$$

If you work this correctly, you should wind up with a quadratic equation to solve.



EDIT: As you pointed out, a quadratic will often have two possible results. This is where that sanity testing I mentioned may come into play.

If you have arranged your equations so that you expect positive results, then it is easy to exclude the negative one, if there is both a negative and positive result. The positive one will be the choice. If there are two positive results, then you apply the currents to both the \$\ref{vd}\$ and \$\ref{vs}\$ computations above. One of the positive values for \$I_D\$ won't make sense, considering your voltage rail, \$V_{CC}\$. So you eliminate that one. (If they are both negative, then you didn't arrange your equations to expect positive results and that is a different problem. Go back and double check your work.)