Appendix A - Equations and Inequalities - Page 420: 31

{$2,3$}

$x^{2}-5x+6=0$ $x^{2}-2x-3x+6=0$ $x(x-2)-3(x-2)=0$ $(x-2)(x-3)=0$ $(x-2)=0$ or $(x-3)=0$ $x=2$ or $x=3$ Check: Substituting $x=2$ in the equation, $(2)^{2}-5(2)+6=0$ $4-10+6=0$ $-6+6=0$ $0=0$ Substituting $x=3$ in the equation, $(3)^{2}-5(3)+6=0$ $9-15+6=0$ $-6+6=0$ $0=0$ Both values check and the solution set is {$2,3$}.