1. **State the problem:** Solve the quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$. 2. **Formula used:** The quadratic formula to find roots is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ This formula gives the solutions for any quadratic equation. 3. **Important rules:** - The term under the square root, $b^2 - 4ac$, is called the discriminant. - If the discriminant is positive, there are two real roots. - If it is zero, there is one real root (a repeated root). - If it is negative, the roots are complex (not real). 4. **Example:** Solve $2x^2 + 3x - 2 = 0$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$$ 6. **Apply the quadratic formula:** $$x = \frac{-3 \pm \sqrt{25}}{2 \times 2} = \frac{-3 \pm 5}{4}$$ 7. **Find the two roots:** - For $+$ sign: $$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{\cancel{2}}{\cancel{4}} = \frac{1}{2}$$ - For $-$ sign: $$x = \frac{-3 - 5}{4} = \frac{-8}{4} = \frac{\cancel{-8}}{\cancel{4}} = -2$$ 8. **Final answer:** The solutions are $x = \frac{1}{2}$ and $x = -2$. This method works for any quadratic equation by substituting the values of $a$, $b$, and $c$.