1. **Problem Statement:** Solve the quadratic equation $x^2 - 5x + 6 = 0$ and sketch its graph. 2. **Formula Used:** The quadratic equation $ax^2 + bx + c = 0$ can be solved using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-5$, and $c=6$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1$$ Since $\Delta > 0$, there are two distinct real roots. 4. **Find the roots:** $$x = \frac{-(-5) \pm \sqrt{1}}{2 \times 1} = \frac{5 \pm 1}{2}$$ So, $$x_1 = \frac{5 + 1}{2} = 3$$ $$x_2 = \frac{5 - 1}{2} = 2$$ 5. **Interpretation:** The solutions to the equation are $x=3$ and $x=2$. 6. **Graph sketch:** The quadratic function is $y = x^2 - 5x + 6$. - It is a parabola opening upwards (since $a=1 > 0$). - The roots (x-intercepts) are at $x=2$ and $x=3$. - The vertex can be found using $x = -\frac{b}{2a} = \frac{5}{2} = 2.5$. - Substitute $x=2.5$ into the function: $$y = (2.5)^2 - 5 \times 2.5 + 6 = 6.25 - 12.5 + 6 = -0.25$$ So the vertex is at $(2.5, -0.25)$. **Final answer:** The roots of the quadratic equation are $x=2$ and $x=3$.