Решите неравенство 5x+3x-8>0.

\[5x^{2} + 3x - 8 > 0\]

\[D = 9 + 160 = 169\]

\[x_{1} = \frac{- 3 + 13}{10} = 1;\ \ \ \]

\[x_{2} = \frac{- 3 - 13}{10} = - 1,6.\]

\[5 \cdot (x + 1,6)(x - 1) > 0\]

\[Ответ:x \in ( - \infty; - 1,6) \cup (1; + \infty).\]