Question

Ajudinha por favor, resolver do delta
X2-4x-5=0

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-4x-5=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-4~e~c=-5\\\\\Delta=(b)^{2}-4(a)(c)=(-4)^{2}-4(1)(-5)=16-(-20)=36\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-4)-\sqrt{36}}{2(1)}=\frac{4-6}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-4)+\sqrt{36}}{2(1)}=\frac{4+6}{2}=\frac{10}{2}=5\\\\S=\{-1,~5\}$

Answer 2

Resposta:

x^2-4x-5=0

∆=(-4)^2-4*1*(-5)

∆=16+20

∆=36

x'=4-√36/2

x'= 4-6/2

x'= -2/2=-1

x=4+6/2= 10/2=5

S={-1,5}