Concentrations at equililibrium At $25 ^\circ C$, $K=0.090$ for the reaction 
  $$\ce{H2O (g) + Cl2O (g) -&gt; 2HOCl (g)}$$ 
  calculate the concentrations of all species at equilibrium for the case: $\pu{1.0 g} \ce{H2O}$ and $\pu{2.0 g} \ce{Cl2O}$ are mixed in a $\pu{1.0 L}$ flask.


I don't know where I go wrong in my calculations:

We have $\pu{0.0555 M}$ $\ce{H2O}$ and $\pu{0.023 M}$ $\ce{Cl2O}$. So at equilibrium, 
\begin{align}
  \ce{[H2O]} &amp;= 0.0555 - x,\\
  \ce{[Cl2O]} &amp;= 0.023 - x\\
  \ce{[HOCl]} &amp;= 2x\\
\therefore K &amp;= \frac{[\ce{HOCl}]^2}{[\ce{H2O}][\ce{Cl2O}]}
\end{align}

After some moving about I get to 
$$-3.91x^2 - 7.1\times10^{-3}x + 1.15\times10^{-4} = 0$$ 
Where I end up getting $x = -8.74\times10^{-3}$ or $x=8.4\times10^{-3}$.

And this is where it all goes wrong, doing some quick calculations, I see that it does not match what the book says the answer is. So I'm doing something wrong, but I don't know what it is or where it's at.
I suggest you to use this schematic: 
\begin{array}{l*{5}{c}}\hline
  \text{Equation} &amp; 
    \ce{H2O} &amp; \ce{Cl2O} &amp; \ce{&lt;=&gt; 2 HClO}\\[2ex]\hline
  \text{Initial concentration} /\pu{M} &amp; 
    \pu{0.055 mol//L} &amp; \pu{0.023 mol//L} &amp; 0  \\[2ex] 
  \text{Equilibrium variation} /\pu{M} &amp;
    - x &amp; -x &amp; +2x \\[2ex]
  \text{Equilibrium concentration} /\pu{M} &amp; 
    \pu{0.055 M} - x &amp; \pu{0.023 M} - x &amp; 2x \\[2ex] \hline
\end{array}

At this point you have to substitute into the equilibrium equation you mentioned:
\begin{align}
  0.090 &amp;= \frac{(2x)^{2}}{(0.055 - x)(0.023 - x)}\\
  0.090 &amp;= \frac{4x^{2}}{0.001265 - 0.055x - 0.023x + x^2}\\
  4x^2  &amp;= 0.090 (0.001265 - 0.055x - 0.023x + x^2)\\
  4x^2  &amp;= 0.090 (0.001265 - 0.078x + x^2)\\
  4x^2  &amp;= 0.000114 - 0.007x + 0.09x^2\\
  0     &amp;=3.91x^2 + 0.007x - 0.000114\\
\end{align}
So the positive result is:
$$ x_2 = \frac{-0.007 + \sqrt{4.9\cdot10^{-5} + 0.00178}}{7.8}\approx 0.0046$$ 

An then equilibrium concentration are for $\ce{[H2O]} = 0.055 - 0.0046 = 0.050$, $\ce{[Cl2O]} = 0.018$ and $\ce{[HClO]} = 2x = 0.092$. This are my results.