SO3]= 0.05 [SO2]= 0.125 [O2]= 0.02 Kc= 7.5x10^-2 What are the final concentrations?

First I calculated Q, and Q>Kc so the reverse reaction is favoured. I created an ICE table and wrote the expression for Kc to solve for X, but the calculator says there is no solution.

Here is the ice table:

        2SO2  +  O2 -> 2SO3
  I     0.125    0.02  0.05
  C      +2x     +x     -2x
  E    0.125+2x 0.02 +x  0.05-2x

7.5 x 10^-2 = ((0.05-2x)^2) / ((0.02+x)(0.125+2x)^2)

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:

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

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