## Intermediate Algebra (12th Edition)

$x=\left\{ 3,5 \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $x^2-8x+15=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a= 1 , b= -8 , c= 15 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(1)(15)}}{2(1)} \\\\ x=\dfrac{8\pm\sqrt{64-60}}{2} \\\\ x=\dfrac{8\pm\sqrt{4}}{2} \\\\ x=\dfrac{8\pm2}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{8-2}{2} \\\\ x=\dfrac{6}{2} \\\\ x=3 \\\\\text{OR}\\\\ x=\dfrac{8+2}{2} \\\\ x=\dfrac{10}{2} \\\\ x=5 .\end{array} Hence, $x=\left\{ 3,5 \right\} .$