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

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?

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– user15489
Jun 22, 2015 at 7:24
• Apart from this, please consider copying the content relating to your question of the linked power point presentation. This site should be self contained, but links can change and then this question would not help anyone any more. Jun 22, 2015 at 7:34
• This is getting better - could you please format your post according to meta.chemistry.stackexchange.com/questions/86/…
– user15489
Jun 22, 2015 at 10:54
• Wow! That is a much improved edit - my retinas are saved!
– user15489
Jun 22, 2015 at 11:51
• Thank you LordStryker & Martin, I understand it now that LordStryker have helped me edit it, and thank you Martin for further explanation for this problem, so I use fractions instead of decimals, correct? Jun 22, 2015 at 16:44

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 $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.