## Trigonometry (11th Edition) Clone

{$-10,+10$} or {$\pm 10$}
$x^{2}-100=0$ $(x)^{2}-(10)^{2}=0$ $(x+10)(x-10)=0$ $(x+10)=0$ or $(x-10)=0$ $x=-10$ or $x=10$ Check: Substituting $x=-10$ in the equation, $(-10)^{2}-(10)^{2}=0$ $100-100=0$ $0=0$ Substituting $x=10$ in the equation, $(10)^{2}-(10)^{2}=0$ $100-100=0$ $0=0$ Both values check and the solution set is {$-10,+10$} or {$\pm10$}