equilibrium constants?

Question

i keep doing this problem but my answer sheet saws the answer is wrong...how do you solve it?

Bromine and chlorine gases react to produce bromine monochloride according to the equation below. The equilibrium constant, Kc for the reaction is 36.0 under the conditions of the experiment.
Br2(g) + Cl2(g) <=> 2 BrCl(g)
If 0.180 moles of Br2(g) and 0.180 moles of Cl2(g) are introduced into a 3.0 L flask and allowed to reach equilibrium, what is the equilibrium molar concentration of BrCl(g)?

Answer 1

First of all convert into concentrations
[Br2]=0.180/3 =0.060
[Cl2] =0.180/3 = 0.060

.. .. .. .. .. .. Br2 + Cl2 <=> 2BrCl
Initial .. .. 0.060 .. 0.060
React .. .. .. x.. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. 2x
Et equil. 0.060-x. 0.060-x .. 2x

Kc= [BrCl]^2 /[Br2][Cl2] = (2x)^2 / (0.06-x)*(0.06-x) =36 =>
4x^2=36(36*10^-4 -0.12x + x^2) =>
32x^2-4.32x +0.1296 =0

This is as quadratic of the form ax^2+bx+f=0. So you get 2 solutions
x1=(-b+Squareroot(b^2-4af) )/(2a)
x2=(-b-Squareroot(b^2-4af) )/(2a)

so x1=0.09 and x2=0.045
x1 is rejected (it is not possible that the amount of Br2 and Cl2 that reacted is more than what you originally had)
so the correct is x=0.045.

BUT [BrCl]=2x= 2*0.045 =0.090 M

I am sure you just divided by 2 (when calculating the roots) instead of 2*32