Resposta:
Alternativa b) {2;4}
Explicação passo a passo:
[tex]\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-6x+8=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-6~e~c=8\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(1)(8)=36-(32)=4\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-6)-\sqrt{4}}{2(1)}=\frac{6-2}{2}=\frac{4}{2}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-6)+\sqrt{4}}{2(1)}=\frac{6+2}{2}=\frac{8}{2}=4\\\\S=\{2,~4\}[/tex]
