Concentrations at equililibrium
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.