Resposta:
S={x∈|R \ x= -2 ou x=13/5}
Explicação passo a passo:
[tex]\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~5x^{2}-3x-26=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=5{;}~b=-3~e~c=-26\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(5)(-26)=9-(-520)=529\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{529}}{2(5)}=\frac{3-23}{10}=\frac{-20}{10}=-2[/tex][tex]\displaystyle x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{529}}{2(5)}=\frac{3+23}{10}=\frac{26\div2}{10\div2}=\frac{13}{5}\\[/tex]
