Question: User: $x^2 + 6x \geq -5$
1. **State the problem:** Solve the inequality $$x^2 + 6x \geq -5$$.
2. **Rewrite the inequality:** Add 5 to both sides to get all terms on one side:
$$x^2 + 6x + 5 \geq 0$$
3. **Factor the quadratic:**
$$x^2 + 6x + 5 = (x + 1)(x + 5)$$
4. **Analyze the inequality:** We want to find where
$$(x + 1)(x + 5) \geq 0$$
5. **Find critical points:** The roots are $x = -1$ and $x = -5$.
6. **Test intervals:**
- For $x < -5$, pick $x = -6$: $( -6 + 1)( -6 + 5) = (-5)(-1) = 5 > 0$
- For $-5 < x < -1$, pick $x = -3$: $( -3 + 1)( -3 + 5) = (-2)(2) = -4 < 0$
- For $x > -1$, pick $x = 0$: $(0 + 1)(0 + 5) = (1)(5) = 5 > 0$
7. **Write solution:** The inequality holds where the product is non-negative, so
$$x \leq -5 \quad \text{or} \quad x \geq -1$$
**Final answer:** $$(-\infty, -5] \cup [-1, \infty)$$