## College Algebra (6th Edition)

$x^{2}-2x-15=0$
Solution set is $\{-3,5\}$ $x=-3$ or $x=5$ $x+3=0$ or $x-5 = 0$ If $x+3=0$ or $x-5 = 0$ then $(x+3)(x-5)=0$ $x^{2}+3x-5x-15=0$ $x^{2}-2x-15=0$ This is the required quadratic equation whose solution set is $\{-3,5\}$