Question$+\infty)^{+\infty}$ therefore, the solution set is PRACTCE: Find the solution set of the following inequalitio (1) $x^{2}+9 x+14>0$ (2) $x^{2}+6 x \geq-5$

Studdy Solution
Combine the results to find the intervals where $(x+5)(x+1) \geq 0$.
The inequality $x^{2}+6x+5 \geq 0$ holds for $x \in (-\infty, -5] \cup [-1, +\infty)$.
Solution sets:
1. For $x^{2}+9x+14>0$: $x \in (-\infty, -7) \cup (-2, +\infty)$.
2. For $x^{2}+6x \geq -5$: $x \in (-\infty, -5] \cup [-1, +\infty)$.