1. The problem is to solve the quadratic inequality $$2x^2 \geq 7x + 15$$ in the set of real numbers. 2. First, rewrite the inequality in standard form by moving all terms to one side: $$2x^2 - 7x - 15 \geq 0$$ 3. To solve this inequality, we find the roots of the quadratic equation: $$2x^2 - 7x - 15 = 0$$ 4. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-7$, and $c=-15$. 5. Calculate the discriminant: $$\Delta = (-7)^2 - 4 \times 2 \times (-15) = 49 + 120 = 169$$ 6. Find the roots: $$x = \frac{-(-7) \pm \sqrt{169}}{2 \times 2} = \frac{7 \pm 13}{4}$$ 7. Calculate each root: - $$x_1 = \frac{7 - 13}{4} = \frac{-6}{4} = -\frac{3}{2}$$ - $$x_2 = \frac{7 + 13}{4} = \frac{20}{4} = 5$$ 8. The quadratic opens upwards since $a=2 > 0$, so the parabola is positive outside the roots and negative between them. 9. Therefore, the solution to the inequality $$2x^2 - 7x - 15 \geq 0$$ is: $$x \leq -\frac{3}{2} \quad \text{or} \quad x \geq 5$$ This means all real numbers less than or equal to $-\frac{3}{2}$ or greater than or equal to $5$ satisfy the inequality.