$2 \cdot 63$ If the roots of the equation $x^{2}+(m-1) x+4-m=0$ are both greater than 2, then the range of values for $m$ is
(A) $m \leqslant-6$.
(B) $-6<m \leqslant-5$.
(C) $m<-3$.
(D) $m<-6$ or $-6<m \leqslant-5$.
(China Hubei Province Huanggang Region Junior High School Mathematics Competition, 1989)
[Solution] Let the two roots of the equation be $x_{1}, x_{2}$, then $x_{1}>2$ and $x_{2}>2$.
Let $y=x-2$, i.e., $x=y+2$ substituted into the original equation gives

i.e.,
$$
\begin{array}{c}
(y+2)^{2}+(m-1)(y+2)+4-m=0, \\
y^{2}+(m+3) y+m+6=0 .
\end{array}
$$
$\because$ The two roots of the original equation are both greater than 2, then the two roots of $\circledast$ are both greater than 0. Thus $\left\{\begin{array}{l}\Delta=(m+3)^{2}-4(m+6) \geqslant 0, \\ m+3>0 .\end{array}\right.$ Solving gives $\quad\left\{\begin{array}{l}m \geqslant 3 \text { or } m \leqslant-5, \\ m>-3 .\end{array}\right.$ $\therefore \quad-6<m \leqslant-5$.
Therefore, the answer is $(B)$.