Resposta:
Explicação passo a passo:
[tex]\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-2x-15=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-2~e~c=-15\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-2)^{2}-4(1)(-15)=4-(-60)=64\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-2)-\sqrt{64}}{2(1)}=\frac{2-8}{2}=\frac{-6}{2}=-3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-2)+\sqrt{64}}{2(1)}=\frac{2+8}{2}=\frac{10}{2}=5[/tex]
[tex]\displaystyle S=\{-3,~5\}[/tex]
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Resposta:
Explicação passo a passo:
[tex]\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-2x-15=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-2~e~c=-15\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-2)^{2}-4(1)(-15)=4-(-60)=64\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-2)-\sqrt{64}}{2(1)}=\frac{2-8}{2}=\frac{-6}{2}=-3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-2)+\sqrt{64}}{2(1)}=\frac{2+8}{2}=\frac{10}{2}=5[/tex]
[tex]\displaystyle S=\{-3,~5\}[/tex]