## Introductory Algebra for College Students (7th Edition)

$\text{$x-$intercepts: } \{ 1, 5 \}$
Setting $y=0$ and solving for $x$ using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), then the $x-$intercept/s of the given equation, $y=x^2-6x+5 ,$ is/are \begin{array}{l}\require{cancel} x^2-6x+5=0 \\\\ x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(5)}}{2(1)} \\\\ x=\dfrac{6\pm\sqrt{36-20}}{2} \\\\ x=\dfrac{6\pm\sqrt{16}}{2} \\\\ x=\dfrac{6\pm\sqrt{(4)^2}}{2} \\\\ x=\dfrac{6\pm4}{2} .\end{array} Hence, the $x-$intercepts are \begin{array}{l}\require{cancel} \text{$x-$intercepts: } \{ 1, 5 \} .\end{array}