1. Let's revisit how we found $x = 3$ as a root of the quadratic equation $x^2 - 5x + 6 = 0$. 2. We use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-5$, and $c=6$. 3. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1$$ 4. Since $\Delta = 1 > 0$, there are two real roots. 5. Calculate the first root using the plus sign: $$x_1 = \frac{-(-5) + \sqrt{1}}{2 \times 1} = \frac{5 + 1}{2} = \frac{6}{2} = 3$$ 6. This is how we got $x = 3$ as one of the solutions. 7. The other root is $x = 2$ calculated similarly with the minus sign.