6. As shown in Figure 2, the linear function $x+2 y-6=0$ intersects with the quadratic function $y=x^{2}$ at points $A$ and $B$. If $C$ is a point on the graph of $y=x^{2}$ such that the area of $\triangle A B C$ is $\frac{35}{8}$, then the coordinates of point $C$ cannot be ( ).
(A) $(-1,1)$
(B) $\left(\frac{1}{2}, \frac{1}{4}\right)$
(C) $\left(\frac{-1+\sqrt{89}}{4}, \frac{45-\sqrt{89}}{8}\right)$
(D) $\left(\frac{1-\sqrt{89}}{4}, \frac{45-\sqrt{89}}{8}\right)$
6. D.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(t, t^{2}\right)$.
From $\left\{\begin{array}{l}x+2 y-6=0, \\ y=x^{2},\end{array}\right.$ we solve to get
$(x, y)=(-2,4),\left(\frac{3}{2}, \frac{9}{4}\right)$.
Thus, $|A B|=\sqrt{\left(-2-\frac{3}{2}\right)^{2}+\left(4-\frac{9}{4}\right)^{2}}=\frac{7 \sqrt{5}}{4}$.
Given $S_{\triangle A B C}=\frac{35}{8}$, the distance from point $C$ to $A B$ is $\sqrt{5}$.
Therefore, point $C$ lies on the line
$y=-\frac{1}{2} x+\frac{1}{2}$ or $y=-\frac{1}{2} x+\frac{11}{2}$
Thus, $t^{2}=-\frac{t}{2}+\frac{1}{2}$ or $t^{2}=-\frac{t}{2}+\frac{11}{2}$
$\Rightarrow t=-1, \frac{1}{2}, \frac{-1 \pm \sqrt{89}}{4}$.
Hence, the coordinates of point $C$ are
$$
\begin{array}{l}
(-1,1),\left(\frac{1}{2}, \frac{1}{4}\right), \\
\left(\frac{-1+\sqrt{89}}{4}, \frac{45-\sqrt{89}}{8}\right), \\
\left(\frac{-1-\sqrt{89}}{4}, \frac{45+\sqrt{89}}{8}\right) .
\end{array}
$$