Найдите множество решений неравенства: (x+3)/5-(x^2-4)/8<=1.

\[\frac{x + 3}{5} - \frac{x^{2} - 4}{8} \leq 1\ \ \ \ \ \ | \cdot 40\]

\[8x + 24 - 5x^{2} + 20 \leq 40\]

\[- 5x^{2} + 8x + 4 \leq 0\]

\[- 5x^{2} + 8x + 4 = 0\]

\[D = 64 + 80 = 144\]

\[x_{1} = \frac{- 8 + 12}{- 10} = - \frac{2}{5} = - 0,4\]

\[x_{2} = \frac{- 8 - 12}{- 10} = 2\]