1. **State the problem:** Solve the quadratic equation $x^2 - 6x + 8 = 0$. 2. **Recall the quadratic formula:** For any quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients. 3. **Identify coefficients:** Here, $a = 1$, $b = -6$, and $c = 8$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-6)^2 - 4 \times 1 \times 8 = 36 - 32 = 4$$ Since $\Delta > 0$, there are two distinct real roots. 5. **Apply the quadratic formula:** $$x = \frac{-(-6) \pm \sqrt{4}}{2 \times 1} = \frac{6 \pm 2}{2}$$ 6. **Find the two solutions:** - For the plus sign: $$x = \frac{6 + 2}{2} = \frac{8}{2} = 4$$ - For the minus sign: $$x = \frac{6 - 2}{2} = \frac{4}{2} = 2$$ 7. **Final answer:** The solutions to the equation are $x = 4$ and $x = 2$. These values satisfy the original equation when substituted back.