375. The curve is given by the equation $y=x^{2}+5 x+3$. Determine the angles of inclination of the tangents to the positive direction of the $O x$ axis, drawn to the curve at the points with abscissas $x=-2$ and $x=0$.
Solution. Let's find the derivative: $y^{\prime}=2 x+5$. Denoting the angle of inclination of the tangent at the point with abscissa $x=-2$ by $\alpha$, and at the point with abscissa $x=0$ by $\beta$, we get

$$
\operatorname{tg} \alpha=y_{x=-2}^{\prime}=2(-2)+5=1, \operatorname{tg} \beta=y_{x=0}^{\prime}=2 \cdot 0+5=5
$$

from which $\alpha=\pi / 4$ (or in degree measure $\alpha=45^{\circ}$ ), and $\beta=\operatorname{arctg} 5$. Using trigonometric function tables, we can also find the degree measure of the angle $\beta: \beta \approx 79^{\circ}$.