Reaction of Fe2+ and S2- to form FeS

Consider this reaction:$$\ce{Fe^{2+} (aq) + S^{2-} (aq) <=> FeS (s)}$$

The equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$. If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, the concentration of $\ce{Fe^2+}$ at equilibrium is found to be $A \times 10^{-17}$. Find $A$.

Let $x$ amount of $\ce{Fe^2+}$ is converted to $\ce{FeS}$. Then$$K = \displaystyle {\frac{1}{(0.06 - x)(0.2 - x)}}$$ Which then leads to the quadratic equation$$ x^2 - 0.26x + 0.012 - 6.25 \times 10^{-18} $$Solving this the $A$ I got is not the answer.Answer is $8.93$