Equilibrium Question?

Question

Nitrogen dioxide decomposes according to the following equation, for which Kp = 0.00063 at a given temperature.
2 NO2(g) <=> 2 NO(g) + O2(g)
A pressure of 5.74 atm of NO2 is placed in a container, and allowed to come to equilibrium. What is the equilibrium partial pressure of O2(g)?

Correct Answer; 0.1731

that's the answer i'm given...

Answer 1

.. .. .. .. .. 2 NO2 <=> 2 NO + O2
Initial .. .. .. 5.74
React .. .. .. 2x
Produce .. .. .. .. .. .. .. 2x .. .. .. x
At Equil .. 5.74-2x .. .. .2x .. .. .. x

Kp = P(NO)^2 * P(O2) / P(NO2)^2 =>
(2x)^2*x/(5.74-2x)^2 =6.3*10^-4 =>
4x^3 = 6.3*10^-4 *(5.74^2-2*2*5.74x +4x^2)=>
4x^3 - (2.52*10^-3)x^2 +0.0144648x -0.020756988=0
x1= 0.166371
x2, x3 are not real so they are rejected
so P(O2)=0.1664 atm

I can't see any mistake in my calculations. I don't know why the answer is different than the one they gave you, unless you didn't type it correctly or I am missing a mistake in my solution.
If you put x=0.1731 and do the calculations you get Kp=7.13*10^-4 instead of 6.3*10^-4, so the answer you give as the correct, can't be correct.

{edit} I honestly can't see how that value is the correct result for those numbers (Kp=6.3*10^-4 and PNO2 initial=5.74 atm). If you get the solution that gives that result (e.g. from your teacher or whoever gave you that value) could you please e-mail it to me? I am really curious to see how they calculated it.