Question: Solve the following inequality algebraically.
$x^2 + 11x + 28 > 0$
1. **State the problem:** Solve the inequality $$x^2 + 11x + 28 > 0$$ algebraically.
2. **Recall the formula:** This is a quadratic inequality. We first find the roots of the quadratic equation $$x^2 + 11x + 28 = 0$$ to determine intervals to test.
3. **Find the roots:** Factor the quadratic:
$$x^2 + 11x + 28 = (x + 7)(x + 4) = 0$$
So the roots are $$x = -7$$ and $$x = -4$$.
4. **Analyze intervals:** The roots divide the number line into three intervals:
- $$(-\infty, -7)$$
- $$(-7, -4)$$
- $$(-4, \infty)$$
5. **Test each interval:**
- For $$x < -7$$, pick $$x = -8$$:
$$(-8 + 7)(-8 + 4) = (-1)(-4) = 4 > 0$$, so inequality holds.
- For $$-7 < x < -4$$, pick $$x = -5$$:
$$(-5 + 7)(-5 + 4) = (2)(-1) = -2 < 0$$, inequality does not hold.
- For $$x > -4$$, pick $$x = 0$$:
$$(0 + 7)(0 + 4) = 7 \times 4 = 28 > 0$$, inequality holds.
6. **Write the solution:** The inequality $$x^2 + 11x + 28 > 0$$ holds for $$x \in (-\infty, -7) \cup (-4, \infty)$$.
**Final answer:** $$\boxed{x < -7 \text{ or } x > -4}$$