1. The problem is to solve the equation labeled as "Ex 2". Since the user did not provide the explicit equation, I will assume a common example for "Ex 2" in algebra: solving a quadratic equation. 2. The general form of a quadratic equation is $$ax^2 + bx + c = 0$$ where $a \neq 0$. 3. The formula to find the roots of the quadratic equation is the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 4. Important rules: - The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots. - 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 two complex roots. 5. Since the exact equation is not given, let's consider an example: $$2x^2 - 4x - 6 = 0$$ 6. Identify coefficients: $a=2$, $b=-4$, $c=-6$. 7. Calculate the discriminant: $$\Delta = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$ 8. Since $\Delta = 64 > 0$, there are two distinct real roots. 9. Apply the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$ 10. Calculate each root: - For the plus sign: $$x = \frac{4 + 8}{4} = \frac{12}{4} = 3$$ - For the minus sign: $$x = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$ 11. Final answer: The solutions to the equation are $$x = 3$$ and $$x = -1$$.