Determine state of MOSFET transistor - trouble identifying Q-point
From a comment: -
If you did some more complex maths to find Vg (i.e. solving the quadratic equation after equating Vd to Vg and solving iD from Vcc and Rd) would you be able to tell if Vg was putting the FET in saturation or otherwise?
From another comment: -

I am not sure how to arrive at the quadratic equation you are
mentioning. Can you give me some more help?

I can leave an answer that shows how you find Vgs (2.0853616 volts)

If you expanded the formulas (bearing in mind that \$V_G = V_D\$), you have you'd get this: -
$$V_G^2 - 2V_TV_G + V_T^2 = \dfrac{i_D}{k} = \dfrac{5}{k\cdot R_D} - \dfrac{V_G}{k\cdot R_D}$$
Therefore: -
$$V_G = \dfrac{2V_T - \frac{1}{k\cdot R_D}±\sqrt{\frac{1}{k^2\cdot R_D^2}-\frac{4V_T}{k\cdot R_D}+4V_T^2 - 4V_T^2+\frac{20}{k\cdot R_D}}}{2}$$
$$V_G = \dfrac{2V_T - \frac{1}{k\cdot R_D}±\sqrt{\frac{1}{k^2\cdot R_D^2}-\frac{4V_T}{k\cdot R_D}+\frac{20}{k\cdot R_D}}}{2}$$
If I plug in the values for k (0.0004), \$R_{D1}\$ (1000 ohm) and \$V_T\$ (2 volts) I get: -
$$V_G = \dfrac{4 - 1.25 ±\sqrt{6.25-20+50}}{2} = 3.7603986447 \text{ or -2.2603986447}$$
Clearly only the positive value is applicable for the MOSFETs.
Can you take it from here?