4・8 For what real number $k$ do the equation $4 x^{2}-(2-k) x+k-5=0$ have two negative roots?
[Solution] Let the two roots of the equation be $\alpha, \beta$. According to the problem, the following conditions must be satisfied:
$$\left\{\begin{array}{l}
\alpha+\beta=\frac{2-k}{4}0 \\
(k-2)^{2}-16(k-5) \geqslant 0
\end{array}\right.$$

From (1) $2-k2$.
From (2) $k-5>0$, i.e., $k>5$.
From (3) $k^{2}-20 k+84=(k-14)(k-6) \geqslant 0$,
i.e., $k \geqslant 14$ or $k \leqslant 6$.
In summary, when $k \geqslant 14$ or $5<k \leqslant 6$, the two roots of the equation are both negative.