Equilibrium concentration, where does the 5.00 for iron thiocyanate complex come from? Trial #1:

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

\begin{array}{|r|c|c|c|}\hline 
\mathrm{Initial~}(M)&amp;\phantom{-}6.00 &amp;\phantom{-}10.00&amp;\phantom{-}0.00\\\hline
\mathrm{Change}~(M) &amp; –4.00 &amp; –4.00 &amp;+4.00  \\\hline
\mathrm{Equilibrium}~(M)&amp;\phantom{-}2.00&amp;\phantom{-}6.00&amp;\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?
You are just making a small sign error:
\begin{align}
ax^2 +bx +c &amp;= 0 &amp; x_{(1;2)} &amp;= \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} &amp;= 0 &amp;
 x_{(1;2)} &amp;= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\
&amp;&amp; x_{(1;2)} &amp;= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\
&amp;&amp;           &amp;= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\
&amp;&amp; x_{(1;2)} &amp;= 16; 5
\end{align}

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