$8.4 \quad \frac{(x+3)(5-x)}{2 x-5}>0$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

$8.4 \quad \frac{(x+3)(5-x)}{2 x-5}>0$.
8.4 The roots of the equations $(x+3)(5-x)=0$ and $2 x-5=0$ are the numbers $x_{1}=-3, x_{2}=2.5, x_{3}=5$, which are not solutions to the given inequality. Therefore, on the number line, we mark them with light circles (Fig. 8.4). These points divide the number line into four intervals. It is easy to determine that for $x>5$, the left side of the inequality is negative - we place a "minus" sign to the right of the point 5 and, moving left, alternate the signs "plus" and "minus". Using Fig. 8.4, we conclude that the solution to the given inequality is the union of two intervals $(-\infty,-3)$ and $(2.5 ; 5)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-269.jpg?height=108&width=575&top_left_y=613&top_left_x=305)

Fig. 8.4

Answer: $(-\infty,-3) \cup(2.5 ; 5)$