1. **State the problem:** Solve the quadratic equation using the Bhaskara formula (quadratic formula). 2. **Formula:** For a quadratic equation $ax^2 + bx + c = 0$, the solutions are given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Important rules:** - The term under the square root, $\Delta = b^2 - 4ac$, is called the discriminant. - If $\Delta > 0$, there are two distinct real roots. - If $\Delta = 0$, there is one real root (a repeated root). - If $\Delta < 0$, there are no real roots (complex roots). 4. **Example:** Suppose the quadratic equation is $2x^2 - 4x - 6 = 0$. 5. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 6. Since $\Delta = 64 > 0$, there are two real roots. 7. Apply the formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 8. Calculate each root: $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 9. **Final answer:** The solutions are $x = 3$ and $x = -1$.