If more no is added, bringing its concentration to 0.700 m, what will the final concentration of no be after equilibrium is re-established?

N2 & O2 <=> 2NO 
0.2M & 0.2M <=> 0.4 M initially 

it sounds like they add enough NO, to momentarily get [NO] to 0.7 
but then that shifts in an attem,pt to re-equilibrate: 
N2 & O2 <=> 2NO 
[0.2+X] & [0.2+X] <=<= [0.7- 2X] 

K = [NO]^2 / [N2] [O2] 

4 = [0.7- 2X] ^2 / [0.2+X] [0.2+X] 

4 = (0.49 - 2.8X + 4X^2) / (0.04 + 0.4X + X^2) 

4 (0.04 + 0.4X + X^2) = (0.49 - 2.8X + 4X^2) 

0.16 + 1.6X + 4X^2 = 0.49 - 2.8X + 4X^2 

0.16 + 1.6X = 0.49 - 2.8X 

1.6X & 2.8X = 0.49 - 0.16 

4.4 X = 0.33 

X = 0.075 

so 
final concentration of NO after equilibrium is re-established would be: 
0.7- 2X 
0.7- 2(0.075) 
0.7- 0.15 
0.55 Molar NO 
 

to check the  
N2 & O2 <=> 2NO 

[0.2+X] & [0.2+X] <=<= [0.7- 2X] 

[0.2+0.075] & [0.2+0.075] <=> [0.7- 2(0.075)] 

[0.2+0.075] & [0.2+0.075] <=> [0.7- 0.15] 

[0.275] & [0.275] <=> 0.55 

K = [NO]^2 / [N2] [O2] 

K = [0.55]^2 / [0.275] [0.275] 

K = 0.3025 / 0.075625 

K = 4 
it checks out 
your answer is 
[NO] = 0.55 Molar

What formula  this one  (x_/