Decrease of volume by half in equilibrium - with specific numbers- how to solve? $$\ce{CO (g) + Cl2(g)&lt;=&gt;COCl2}(g)$$
  We have
  2 mols of $\ce{CO}$, 1 Mol of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
  contained in 1 L.
  The system is currently in equilibrium. 
  
  We then decrease the volume by half to 0.5 L
  
  What will be the new concentration in equilibrium of CO?


I tried calculating at first the constant of concentration; 

$Kc=\frac{[\ce{CoCl2}]}{[\ce{CO}]\times[\ce{Cl2}]}$

I found it to be $4.8825$

then I calculated the concentration with the new Volume:

$[\ce{CO}]= \pu{2mol}/\pu{0.5L}= \pu{4 M}$

$[\ce{Cl2}]= \pu{1mol}/\pu{0.5L}= \pu{2 M}$

$[\ce{CoCl2}]= \pu{9.765mol}/\pu{0.5L}= \pu{19.53 M}$

obviously this wasn't in equilibrium so I went to do this:

$\ce{CO + Cl2&lt;=&gt;COCl2}$

$4M-x$, $2M-x$,  $19.53+x$

(as the number of mols in the products is smaller and the system wants to fix the increase in pressure) 

When I calculated all of this in the Kc formula it came to me that the Concentration of CO increased, and is now larger than 2 M , 
and in the textbook answers it is stated that it is supposed to be less than 2M, how is it possible? 

I would be very glad for help with calculations and technique of solving as well!
I solved the quadratic equation by Wolfram Alpha



The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}&lt;\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}&lt;\pu{2M}$, if we have
1 mol of $\ce{CO}$, 2 Mols of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
contained in 1 L.