1. In the set of real numbers, solve the inequality

$$
\frac{4 x+15-4 x^{2}}{\sqrt{4 x+15}+2 x} \geq 0
$$
Solution. We have:

$$
\frac{4 x+15-4 x^{2}}{\sqrt{4 x+15}+2 x} \geq 0 \Leftrightarrow \frac{(\sqrt{4 x+15}+2 x)(\sqrt{4 x+15}-2 x)}{\sqrt{4 x+15}+2 x} \geq 0 \Leftrightarrow\left\{\begin{array}{l}
\sqrt{4 x+15}+2 x \neq 0 \\
\sqrt{4 x+15}-2 x \geq 0
\end{array}\right.
$$

Since $\sqrt{4 x+15}+2 x=0$ for $x=-\frac{3}{2}$, and

![](https://cdn.mathpix.com/cropped/2024_06_05_6b1323df06a0e4d4e36cg-3.jpg?height=257&width=1036&top_left_y=1530&top_left_x=352)

Finally, $x \in\left[-\frac{15}{4},-\frac{3}{2}\right) \cup\left(-\frac{3}{2}, \frac{5}{2}\right]$.