1. The problem is to solve the quadratic equation $x^2 - 3x - 10 = 0$. 2. The general form of a quadratic equation is $ax^2 + bx + c = 0$. 3. We use the quadratic formula to find the roots: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-3$, and $c=-10$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-3)^2 - 4 \times 1 \times (-10) = 9 + 40 = 49$$ 5. Since $\Delta > 0$, there are two distinct real roots. 6. Substitute values into the quadratic formula: $$x = \frac{-(-3) \pm \sqrt{49}}{2 \times 1} = \frac{3 \pm 7}{2}$$ 7. Calculate each root: - For the plus sign: $$x = \frac{3 + 7}{2} = \frac{10}{2} = 5$$ - For the minus sign: $$x = \frac{3 - 7}{2} = \frac{\cancel{3} - 7}{\cancel{2}} = \frac{-4}{2} = -2$$ 8. Therefore, the solutions to the equation $x^2 - 3x - 10 = 0$ are: $$x = 5 \quad \text{or} \quad x = -2$$