Appendix A - Equations and Inequalities - Page 417: 34

{$2,-4$}

$x^{2}+2x-8=0$ $x^{2}-2x+4x-8=0$ $x(x-2)+4(x-2)=0$ $(x-2)(x+4)=0$ $(x-2)=0$ or $(x+4)=0$ $x=2$ or $x=-4$ Check: Substituting $x=-4$ in the equation, $(2)^{2}+2(2)-8=0$ $4+4-8=0$ $8-8=0$ $0=0$ Substituting $x=3$ in the equation, $(-4)^{2}+2(-4)-8=0$ $16-8-8=0$ $16-16=0$ $0=0$ Both values check and the solution set is {$2,-4$}