1. Let's solve a random algebra problem: Find the roots of the quadratic equation $$x^2 - 5x + 6 = 0$$. 2. The formula to find roots of a quadratic equation $$ax^2 + bx + c = 0$$ is given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. Here, $$a = 1$$, $$b = -5$$, and $$c = 6$$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1$$ 5. Since $$\Delta > 0$$, there are two distinct real roots. 6. Calculate the roots: $$x_1 = \frac{-(-5) + \sqrt{1}}{2 \times 1} = \frac{5 + 1}{2} = 3$$ $$x_2 = \frac{-(-5) - \sqrt{1}}{2 \times 1} = \frac{5 - 1}{2} = 2$$ 7. Therefore, the solutions to the equation $$x^2 - 5x + 6 = 0$$ are $$x = 3$$ and $$x = 2$$.