Equilibrium Bond Dissociation
Your values for $x$ are perfectly correct and lead to the correct concentrations, though I can understand why you are perturbed by negative numbers. Let's see what's going on.

First, remember that $x$ is not the final concentration. It has to do with the change in concentration. If $x$ is negative, then that means the concentration is changing in the opposite direction than what you thought it was. We'll come back to that.

Let's take one of those values of $x$ and plug it into your equilibrium concentration expressions. Let's start with $x=-0.167$.

For $\ce{N2O4}$:

$$[\ce{N2O4}]=\dfrac{1.00-x}{4.00}=\dfrac{1.00--0.167}{4.00}=\dfrac{1.167}{4.00}=0.292$$

For $\ce{NO2}$

$$[\ce{NO2}]=\dfrac{0.500+2x}{4.00}=\dfrac{0.500+2(-0.167)}{4.00}=\dfrac{0.164}{4.00}=0.041$$

Notice anything? The first concentartion is spot on with one of the answers, and the second is probably just a rounding error somewhere (yours or mine).


  Now about that negative value for $x$.


Notice that you have the concentration of $\ce{N2O4}$ decreasing in your ICE table, but the final concentration is higher than the initial (0.292 vs 0.250). Your assumption about the direction of the shift in the reaction toward was not correct. Not all reactions shift toward the right. How could we have known that the reaction would have shifted toward the left? We could have calculated the reaction quotient $Q_c$ at the initial conditions. If $Q_c &gt; K_c $, then the products are in excess and the reaction shifts to the left. If $Q_c &lt; K_c$, then the reactants are in excess and the reaction shifts to the right. We calculate $Q_c$ by putting the initial concentrations into the law of mass action.

$$Q_c =\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}=\dfrac{\left(\dfrac{0.500}{4.00}\right)}{\dfrac{1.00}{4.00}}=\dfrac{0.125^2}{0.250}=6.25\times 10^{-2}&gt;5.9\times 10^{-3}$$

Thus, your ICE table should have the signs of $x$ reversed: the concentration of $\ce{N2O4}$ is increasing and the concentration of $\ce{NO2}$ is decreasing.

$$\begin{array}{|c|c|c|} \hline
\ &amp; \ce{N2O4} &amp; \ce{NO2} \\ \hline
I &amp; \dfrac{1.00}{4.00} &amp; \dfrac{0.500}{4}\\ \hline
C &amp; \dfrac{+x}{4.00} &amp; \dfrac{-2x}{4.00} \\ \hline
E &amp; \dfrac{1.00+x}{4.00} &amp; \dfrac{0.500-2x}{4}\\ \hline
\end{array}$$

$$5.9\times 10^{-3}=\dfrac{\left(\dfrac{0.5.00-2x}{4.00}\right)^2}{\dfrac{1.00+x}{4.00}}$$

Now $x=+0.167$ and $x=+0.338$. $x=+0.167$ leads to the same answer since the signs are switched everywhere.

The other value $x=\pm 0.338$, whether negative or positive, is not valid since $0.500-2(.338)&lt;0$.