Question

x^2+x-12=0 Alguem pode me ajudar ?

Answer 1

x²+x-12=0

x=-b+-\/b²-4ac/2a

x=-(1)+-\/(1)²-4(1)(-12)/2(1)

x=-1+-\/1+48/2

x=-1+-\/49/2

x=-1+-7/2

x'=-1+7/2

x'=6/2

x'=3

x''=-1-7/2

x''=-8/2

x"=-4

S={-4,3}

Answer 2

Resposta:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+x-12=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=1~e~c=-12\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-12)=1-(-48)=49\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{49}}{2(1)}=\frac{-1-7}{2}=\frac{-8}{2}=-4\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{49}}{2(1)}=\frac{-1+7}{2}=\frac{6}{2}=3\\\\S=\{-4,~3\}$