1. **State the problem:** Solve the quadratic equation $x^2 - 6x + 5 = 0$. 2. **Formula and rules:** The quadratic formula to solve $ax^2 + bx + c = 0$ is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-6$, and $c=5$. 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. **Find the roots:** $$x = \frac{-(-6) \pm \sqrt{16}}{2 \times 1} = \frac{6 \pm 4}{2}$$ 5. **Evaluate each root:** - For $+$ sign: $$x = \frac{6 + 4}{2} = \frac{10}{2} = 5$$ - For $-$ sign: $$x = \frac{6 - 4}{2} = \frac{2}{2} = 1$$ **Final answer:** The solutions to the equation are $x = 5$ and $x = 1$.