6. If the inequality $x^{2}+|x-a|<2$ has at least one negative solution, then the range of the parameter $a$ is ( ).
(A) $\left(-\frac{5}{4}, 2\right)$
(B) $\left(-\frac{7}{4}, 2\right)$
(C) $\left(-\frac{9}{4}, 2\right)$
(D) $\left(-\frac{7}{4}, 3\right)$
6.C.

From the original inequality, we know
$$
\begin{array}{l}
|x-a|<2-x^{2} \\
\Rightarrow x^{2}-2<x-a<2-x^{2} .
\end{array}
$$

As shown in Figure 4, the curve segments $C_{1}$
and $C_{2}$ are the parts of the parabola $y=2-x^{2}$ above the x-axis and the parabola $y=x^{2}-2$ below the x-axis, respectively. When equation (1) has at least one negative solution, the line $y=x-a$ has two critical positions: one is passing through the intersection point $(0,-2)$ of $C_{2}$ and the y-axis, at which time $a=2$; the other is when it is tangent to $C_{1}$.
For the latter, from $x-a=2-x^{2}$, we have
$$
\begin{array}{l}
x^{2}+x-a-2=0 \\
\Rightarrow \Delta=1+4(a+2)=0 \Rightarrow a=-\frac{9}{4} .
\end{array}
$$

Therefore, when $-\frac{9}{4}<a<2$, the original inequality has at least one negative solution.