1. **State the problem:** Solve the inequality $2x^2 + 3x - 5 \leq 0$. 2. **Formula and rules:** To solve a quadratic inequality, first find the roots of the quadratic equation $2x^2 + 3x - 5 = 0$ by using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=3$, and $c=-5$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times (-5) = 9 + 40 = 49$$ 4. **Find the roots:** $$x = \frac{-3 \pm \sqrt{49}}{2 \times 2} = \frac{-3 \pm 7}{4}$$ 5. **Evaluate each root:** - For $+$ sign: $$x = \frac{-3 + 7}{4} = \frac{4}{4} = 1$$ - For $-$ sign: $$x = \frac{-3 - 7}{4} = \frac{-10}{4} = -\frac{5}{2}$$ 6. **Determine intervals:** The roots split the number line into three intervals: $$(-\infty, -\frac{5}{2}), \quad [-\frac{5}{2}, 1], \quad (1, \infty)$$ 7. **Test the sign of the quadratic in each interval:** - For $x = -3$ (in $(-\infty, -\frac{5}{2})$): $$2(-3)^2 + 3(-3) - 5 = 18 - 9 - 5 = 4 > 0$$ - For $x = 0$ (in $(-\frac{5}{2}, 1)$): $$2(0)^2 + 3(0) - 5 = -5 < 0$$ - For $x = 2$ (in $(1, \infty)$): $$2(2)^2 + 3(2) - 5 = 8 + 6 - 5 = 9 > 0$$ 8. **Conclusion:** The quadratic is less than or equal to zero between the roots, including the roots themselves: $$\boxed{-\frac{5}{2} \leq x \leq 1}$$