Find Q-Point for a Transistor Circuit
First use 0.783 mA and find Vgs 

Vgs = 5 - 0.783*2.7 = 2.8859V 

Next we we find Vgs for Id = 1.57mA 

Vgs = 5 - 1.57*2.7 = 0.761V  

As you can see only Vgs = 2.886V and Id = 0.783mA is valid because Vgs cannot be lower than Vt. For Vgs &lt; Vt the MOSFET is in cut-off region. 

So the next time try to solve for Vgs directly 

$$ I_{D} = K_n(V_{GS} - V_{TN})^2$$
$$ I_{D} = \frac{V_g - V_{GS}}{Rs}$$ 
$$ \frac{V_g - V_{GS}}{Rs}= K_n(V_{GS} - V_{TN})^2$$
$$ \frac{5V - V_{GS}}{2.7k\Omega}=1 {mA}/{V^2}(V_{GS} - 2V)^2$$

And the solution is 

$$V_{GS}=0.744579V ;V_{GS}=2.88505V$$

As for this Id = 1.11mA let as see what we get

Vgs = 5-1.11*2.7 = 2.003V so for sure this is not the valid answer.