1. The problem is to solve the quadratic equation $2x^2 - 4x - 6 = 0$. 2. The formula to solve quadratic equations is the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the equation $ax^2 + bx + c = 0$. 3. Here, $a=2$, $b=-4$, and $c=-6$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$. 5. Since $\Delta > 0$, there are two real solutions. 6. Substitute values into the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 7. Calculate each solution: $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 8. Therefore, the solutions to the equation $2x^2 - 4x - 6 = 0$ are $x=3$ and $x=-1$.