One, (20 points) A certain musical instrument has 10 holes, sequentially labeled as hole 1, hole 2, $\cdots \cdots$, hole 10. When played, the sound quality index $D$ of the $n$-th hole is given by the relationship
$$
D=n^{2}+k n+90 .
$$

The lowest sound quality index of the instrument is $4 k+106$. Find the range of values for the constant $k$.


For the parabola $D=n^{2}+k n+90$, the axis of symmetry is $n=-\frac{k}{2}$.
(1) When $-\frac{k}{2} \leqslant 1$, i.e., $k \geqslant-2$, we have
$$
n=1, D=4 k+106 \text {. }
$$

Thus, $1^{2}+k+90=4 k+106$.
Solving this, we get $k=-5$ (which does not meet the condition).
(2) When $-\frac{k}{2} \geqslant 10$, i.e., $k \leqslant-20$, we have
$$
n=10, D=4 k+106 \text {. }
$$

Thus, $10^{2}+10 k+90=4 k+106$.
Solving this, we get $k=-14$ (which does not meet the condition).
(3) When $1<-\frac{k}{2}<10$, i.e., $-20<k<-2$, $n$ is within the range $-\frac{k}{2}-\frac{1}{2} \leqslant n \leqslant-\frac{k}{2}+\frac{1}{2}$, and $D$ has the lowest value, which is $4 k+106$.
$$
\text { Thus }\left(-\frac{k}{2}+\frac{1}{2}\right)^{2}+k\left(-\frac{k}{2}+\frac{1}{2}\right)+90 \geqslant 4 k+106 \text {. }
$$

Simplifying, we get $(k+7)(k+9) \leqslant 0$.
Solving this, we get $-9 \leqslant k \leqslant-7$.
In conclusion, the range of $k$ is $-9 \leqslant k \leqslant-7$.