Equilibrium concentration, where does the 5.00 for iron thiocyanate complex come from?

Question

Trial #1:

$$\ce{Fe^3+(aq) + SCN^- (aq) -> FeSCN^2+(aq)}$$

\begin{array}{|r|c|c|c|}\hline \mathrm{Initial~}(M)&\phantom{-}6.00 &\phantom{-}10.00&\phantom{-}0.00\\\hline \mathrm{Change}~(M) & –4.00 & –4.00 &+4.00 \\\hline \mathrm{Equilibrium}~(M)&\phantom{-}2.00&\phantom{-}6.00&\phantom{-}4.00\\\hline \end{array} $$ K = 0.333$$

Consider the reaction represented by the equation: $$\ce{Fe^3+(aq) + SCN- (aq) \longrightarrow [Fe(SCN)]^2+(aq)}$$

Trial #2:

Initial: $10.0 ~M\ce{~Fe^3+(aq)}$ and $8.00~M\ce{ ~SCN- (aq)}$ (same temperature as Trial #1)

Equilibrium: $x~M\ce{~FeSCN^2+(aq)}$ 

I have tried putting $10.0$ and $8.00$ in an ice table, then when I have finished I start solving to get my equation as $$0=\frac{0.33x}{(10.0-x)(8.00-x)}$$ when I'm done I come out with $$0.33=\frac{x}{(x^2-18x+80)}$$ then I plug in for quadratic equation, $$x=\frac{-6.94\pm\sqrt{6.94^2-4\times0.33\times26.4}}{2\times0.33}$$ my answers came out to be $-5$ and $-16.03$. What am I doing wrong?

Answer 1

You are just making a small sign error: \begin{align} ax^2 +bx +c &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}b \pm\sqrt{b^2 - 4ac}}{2a} \end{align}

Taking $K=\frac13$: \begin{align} \frac13x^2 - \frac{21}{3}x +\frac{80}{3} &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\ && x_{(1;2)} &= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\ && &= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\ && x_{(1;2)} &= 16; 5 \end{align}

The first value obviously does not make any sense, hence $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.