Chemistry Equilibrium using ice and solving for x using the quadratic equation

Chemistry Equilibrium using ice and solving for x using the quadratic equation?
Following reaction has an equilibrium constant , Kc, of 12 @ 900*C
CO(g)+Cl2(g)<--->COCl2(g)
Initially .57 moles Cl2(g) and .57 moles CO(g) in a 1L container. How many moles of CO will be present when system reaches equilibrium?

a) .02 moles
b) .18 moles
c) .04 moles
d) .5 moles

I understand how to formulate the ICE table im having problems with the math particularly right after the ice table.

1st step: gather initial info
CO= .57, Cl2=.57
1l container
step 2: change to molarity
(changed for CO and Cl2) .57moles/1L= .57mol/L

step 3: ICE table

CO(g)+Cl2(g)<--->COCl2(g)
I .57M .57M 0
C -x -x +x
E .57-x .57-x x

Step 4:

Kc= 12=[COCl2])/[CO][Cl2]

12=x/(.57-x)^2 <---- after this step I'm pretty much lost. My math skills evade me not sure how to get x by itself and use the quadratic equation... PLEASE HELP ME OUT WITH A STEP BY STEP AFTER THIS. PLEASE CORRECT ANY OF MY MISTAKES. HELP ON HOW TO GET X BY ITSELF AND USING THE QUADRATIC EQUATION. the answers that i have been getting are out there. i know somewhere in the end i have to turn the answer into moles but i cant even get that far....

It looks like everything you've done to this point is correct. So,

12=x/(.57-x)^2

First, square the term in the denominator:

12 = x / (0.3249 - 1.14 x + x^2)

Now, cross multiply:

3.90 -13.68x + 12 x^2 = x

Finally subtract x from both sides:
3.90 - 14.68 x + 12x^2 = 0

Now, use the quadratic equation to calculate x:

x = [-b +/- Square root (b^2 - 4ac)] / 2a

x = 14.68 +/- square root (14.68^2 - 4(12)3.90) / 24

x = 14.68 +/- square root (215.5 - 187.2) / 24
x = 14.68 +/- 5.32 / 24

x = 0.83 or x = 0.39

X cannot be 0.83, so x = 0.39

Therefore, remaining concentration of CO = 0.57 - 0.39 = 0.18 M x 1 L = 0.18 moles CO

Hope that all helps and makes sense.