How to calculate the concentration of NO at equilibrium? $\ce{N2(g) + O2(g) &lt;=&gt; 2NO(g)}$, $K_c = 4.1 \cdot 10^{-4} (T=2000~^\circ\mathrm{C})$
  What is $\ce{[NO]}$ when a mixture of $0.20~\mathrm{mol}$ $\ce{N2(g)}$ and $0.15~\mathrm{mol}$ $\ce{O2(g)}$ reach equilibrium in a $1.0~\mathrm{L}$ container at $T=2000~^\circ\mathrm{C}$?


When K is $\cdot10^{-4}$, you can assume the change is negligible, and just using the concentrations you can get from the given mole amount and the container volume, I get the right answer. When plugging back into, to see if it matches the give $K_c$, it's right:
$$((0.20~\mathrm{M}~\ce{N2})(0.15~\mathrm{M}~\ce{O2}) \cdot 4.1\cdot10^{-4} K_c) = x^2.$$ Ans: $\ce{[NO]} = 0.0035~\mathrm{M}$. 

From what I've read and heard you should be able to do it 'the hard' way, using the quadratic formula, like you would if $K_c$ wasn't to $10^{-4}$, and small enough. But I can't seem to get the right answer trying to do it that way.

Rearranging everything, the equation I get to plug into the quadratic formula is
$$x^2 - 0.35x + 0.03,$$ then I get $0.2$ and $0.15$ as the answers. $0.2$ is too high, subtracting it from the $0.15~\mathrm{M}$ $\ce{O2}$ initial wouldn't be right. So using the $4.15$ as $x$, the change in $\mathrm{M}$, and subtraction that from the initial concentrations I have, then plugging that into the equation for $K_c$, even tried rearranging it in case I have got something mixed up, but I don't get anything close to the given $K_c$.
$K_c = \frac{[\ce{NO}]^2}{[\ce{N2}][\ce{O2}]}$

Let x = [NO]

$\frac{x^2}{(0.20M-0.5x)(0.15M-0.5x)} = 0.00041$

$x^2 = 0.00041(0.25x^2 -0.175xM + 0.030M^2)$

$x^2 + 0.000072xM -0.0000123M^2 = 0$