Resposta:
Explicação passo a passo:
[tex]\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-6x-16=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~a=1{;}~b=-6~e~c=-16\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(1)(-16)=36-(-64)=100\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-6)-\sqrt{100}}{2(1)}=\frac{6-10}{2}=\frac{-4}{2}=-2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-6)+\sqrt{100}}{2(1)}=\frac{6+10}{2}=\frac{16}{2}=8\\\\S=\{-2,8\}[/tex]
