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

$\text{$x-$intercepts: } \{ 6, 2 \}$
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, $y=-x^2+8x-12 ,$ is/are \begin{array}{l}\require{cancel} -x^2+8x-12=0 \\\\ x=\dfrac{-8\pm\sqrt{8^2-4(-1)(-12)}}{2(-1)} \\\\ x=\dfrac{-8\pm\sqrt{64-48}}{-2} \\\\ x=\dfrac{-8\pm\sqrt{16}}{-2} \\\\ x=\dfrac{-8\pm\sqrt{(4)^2}}{-2} \\\\ x=\dfrac{-8\pm4}{-2} .\end{array} Hence, the $x-$intercepts are \begin{array}{l}\require{cancel} \text{$x-$intercepts: } \{ 6, 2 \} .\end{array}