To solve the inequality $x^2 - 3x - 10 \geq 0$, we can follow these steps:

1. Rewrite the inequality as an equation:
   Solve for $x$ in the equation: $x^2 - 3x - 10 = 0$ Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = 1$, $b = -3$, and $c = -10$:

   $x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2}$

   So, the solutions are: $x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-4}{2} = -2$

2. Analyze intervals:
   These solutions, $x = -2$ and $x = 5$, divide the real number line into three intervals:

   • $x < -2$
   • $-2 \leq x \leq 5$
   • $x > 5$

3. Test each interval:
   Choose a test point in each interval and substitute it back into the inequality $x^2 - 3x - 10 \geq 0$:

   • For $x < -2$ (e.g., $x = -3$): $(-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 \quad (\text{which is } \geq 0)$
   • For $-2 < x < 5$ (e.g., $x = 0$): $0^2 - 3(0) - 10 = -10 \quad (\text{which is } < 0)$
   • For $x > 5$ (e.g., $x = 6$): $6^2 - 3(6) - 10 = 36 - 18 - 10 = 8 \quad (\text{which is } \geq 0)$

4. Conclusion:
   The inequality $x^2 - 3x - 10 \geq 0$ holds for $x \in (-\infty, -2] \cup [5, \infty)$.

Final Answer:

$x \in (-\infty, -2] \cup [5, \infty)$

Would you like further clarification on any of these steps?

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Related questions:

1. How would you solve a similar inequality with different coefficients, like $x^2 - 4x + 3 \geq 0$?
2. What is the difference between solving $x^2 - 3x - 10 \leq 0$ and $x^2 - 3x - 10 > 0$?
3. How do we know that substituting values into intervals accurately determines the solution?
4. What if we had a strict inequality $x^2 - 3x - 10 > 0$? How would the solution set change?
5. How would you graph the function $y = x^2 - 3x - 10$ and indicate the regions where $y \geq 0$?

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Tip: When analyzing quadratic inequalities, remember that the roots divide the number line into intervals that you can test to determine where the inequality holds.