# ice tables equilibirium q

I was given initial concentrations for the following reaction:

$$\ce{2SO2 + O2 <=> 2SO3}\ \ \ K_c =7.5\times 10^{-2}\\ [\ce{SO3}]= 0.05\ \mathrm{M}\\ [\ce{SO2}]= 0.125\ \mathrm{M}\\ [\ce{O2}]= 0.02\ \mathrm{M}$$

What are the final concentrations?

First I calculated $Q$, and found that $Q>K_c$ so the reverse reaction is favoured.

I created an ICE table and wrote the expression for $K_c$ to solve for $x$, but the calculator says there is no solution.

Here is the ice table:

$$\begin{array}{|c|c|c|c|} \hline \ & [\ce{SO2}] & [\ce{O2}] & [\ce{SO3}]\\ \hline I & 0.125 & 0.02 & 0.05 \\ C & +2x & +x & -2x \\ E & 0.125+2x & 0.02+x & 0.05-2x\\ \hline \end{array}\\$$

$$7.5 \times 10^{-2} = \dfrac{(0.05-2x)^2}{(0.02+x)(0.125+2x)^2}$$

You have done everything correctly so far. Your calculator might not be able to factor the expression as it is written. It is a cubic equation with three roots and two discontinuous points. You may be used to ICE-type equilibrium problems where you can simplify parts of the expression because $x$ is small compared to the initial concentrations. This is not the case in this problem. Factoring the cubic polynomial is not always easy.
Your mass action expression can be rearranged to the following cubic equation so long as $x \ne -0.02$ and $x \ne -0.0625$, both of which are nonsense anyway. If your calculator cannot factor this expression, try using Wolfram Alpha.
$$0.3 x^3-3.9565 x^2+0.201922 x-0.00247656 = 0$$