5. For the equation $x^{2}+(m-2) x+5-m=0$, if both roots are greater than $2$, the range of $m$ is ( ).
A. $(-5,-4]$
B. $(-\infty,-4]$
C. $(-\infty,-2)$
D. $(-\infty,-5) \cup(-5,-4]$
5. A. From the problem, we get $\left\{\begin{array}{l}\Delta=(m-2)^{2}-4(5-m) \geqslant 0 \\ -\frac{m-2}{2}>2 \\ f(2)=4+2(m-2)+5-m>0,\end{array}\right.$ solving this yields $-5<m \leqslant-4$.