1. **State the problem:** Solve the quadratic equation $x^2 - 6x + 5 = 0$. 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 $a=1$, $b=-6$, and $c=5$ in this case. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-6)^2 - 4 \times 1 \times 5 = 36 - 20 = 16$$ Since $\Delta > 0$, there are two distinct real roots. 4. **Apply the quadratic formula:** $$x = \frac{-(-6) \pm \sqrt{16}}{2 \times 1} = \frac{6 \pm 4}{2}$$ 5. **Find the two solutions:** - For the plus sign: $$x = \frac{6 + 4}{2} = \frac{10}{2} = 5$$ - For the minus sign: $$x = \frac{6 - 4}{2} = \frac{2}{2} = 1$$ 6. **Final answer:** The solutions to the equation are $x=5$ and $x=1$.