1. **State the problem:** Solve the quadratic equation $2x^2 + 3x - 2 = 0$ using the discriminant formula $D = b^2 - 4ac$. 2. **Identify coefficients:** Here, $a = 2$, $b = 3$, and $c = -2$. 3. **Calculate the discriminant:** $$D = b^2 - 4ac = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$$ 4. **Interpret the discriminant:** Since $D > 0$, there are two distinct real roots. 5. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{D}}{2a}$$ 6. **Substitute values:** $$x = \frac{-3 \pm \sqrt{25}}{2 \times 2} = \frac{-3 \pm 5}{4}$$ 7. **Calculate each root:** - For $+$ sign: $$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{\cancel{2}}{\cancel{4}} = \frac{1}{2}$$ - For $-$ sign: $$x = \frac{-3 - 5}{4} = \frac{-8}{4} = \frac{\cancel{-8}}{\cancel{4}} = -2$$ **Final answer:** The solutions are $x = \frac{1}{2}$ and $x = -2$.