1. **Problem statement:** Find the value of $\alpha - \beta$ where $\alpha$ and $\beta$ are roots of the quadratic equation $$3x^2 - 10x - 8 = 0$$ with $\alpha > \beta$. 2. **Formula and rules:** For a quadratic equation $ax^2 + bx + c = 0$, the roots $\alpha$ and $\beta$ satisfy: - Sum of roots: $\alpha + \beta = -\frac{b}{a}$ - Product of roots: $\alpha \beta = \frac{c}{a}$ The difference between roots is given by: $$\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta}$$ 3. **Calculate sum and product:** - $a = 3$, $b = -10$, $c = -8$ - Sum: $\alpha + \beta = -\frac{-10}{3} = \frac{10}{3}$ - Product: $\alpha \beta = \frac{-8}{3} = -\frac{8}{3}$ 4. **Calculate difference:** $$\alpha - \beta = \sqrt{\left(\frac{10}{3}\right)^2 - 4 \times \left(-\frac{8}{3}\right)} = \sqrt{\frac{100}{9} + \frac{32}{3}}$$ Convert $\frac{32}{3}$ to ninths: $$\frac{32}{3} = \frac{96}{9}$$ So, $$\alpha - \beta = \sqrt{\frac{100}{9} + \frac{96}{9}} = \sqrt{\frac{196}{9}} = \frac{14}{3}$$ 5. **Answer:** The value of $\alpha - \beta$ is $\frac{14}{3}$.