Resposta:
Explicação passo-a-passo:
[tex]\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-5x+6=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-5~e~c=6\\\\C\'alculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-5)^{2}-4(1)(6)=25-(24)=1\\\\C\'alculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-5)-\sqrt{1}}{2(1)}=\frac{5-1}{2}=\frac{4}{2}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-5)+\sqrt{1}}{2(1)}=\frac{5+1}{2}=\frac{6}{2}=3\\\\S=\{2,~3\}[/tex]
