1. **State the problem:** Solve the inequality $$-15x \leq 2x^2 + 18$$. 2. **Rewrite the inequality:** Move all terms to one side to set the inequality to zero: $$0 \leq 2x^2 + 18 + 15x$$ which simplifies to $$0 \leq 2x^2 + 15x + 18$$. 3. **Analyze the quadratic inequality:** We want to find where $$2x^2 + 15x + 18 \geq 0$$. 4. **Find the roots of the quadratic equation:** Solve $$2x^2 + 15x + 18 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=15$, and $c=18$. Calculate the discriminant: $$\Delta = 15^2 - 4 \times 2 \times 18 = 225 - 144 = 81$$. Calculate the roots: $$x = \frac{-15 \pm \sqrt{81}}{2 \times 2} = \frac{-15 \pm 9}{4}$$. So, $$x_1 = \frac{-15 - 9}{4} = \frac{-24}{4} = -6$$ $$x_2 = \frac{-15 + 9}{4} = \frac{-6}{4} = -\frac{3}{2}$$. 5. **Determine the intervals where the quadratic is positive or zero:** Since $a=2 > 0$, the parabola opens upwards. - The quadratic is positive outside the roots: $(-\infty, -6)$ and $(-\frac{3}{2}, \infty)$. - The quadratic is zero at $x = -6$ and $x = -\frac{3}{2}$. - The quadratic is negative between the roots: $(-6, -\frac{3}{2})$. 6. **Write the solution to the inequality:** $$x \in (-\infty, -6] \cup \left[-\frac{3}{2}, \infty\right)$$. **Final answer:** $$\boxed{x \leq -6 \text{ or } x \geq -\frac{3}{2}}$$