Cadium metal is added to 0.350 L of an aqueous solution in which [Cr^3+] = 1.00 M

What are the concentrations of the different ionic species at equilibrium?

 

Chem Equation:

 2 Cr^3+ (aq) + Cd (s) <--> 2 Cr^2+ (aq) + Cd^2+ (aq)

Kc = 0.288

 

Answer:

So I set up an ICE chart

     2 Cr^3+ (aq) + Cd (s) <--> 2 Cr^2+ (aq) + Cd^2+ (aq)

 I    1.00 M                             

C    -2x M                                  + 2x                 + x

E   (1.00 - x) M                             2x                   x

 

Then sub that into the Kc equation:

0.288 = (2x)^2 (x)  divided by (1.00 - 2x)^2

 

So now I don't know how to solve for x.  I cannot use the quadractic formula. 

Thanks for your help!