Respostas

2014-07-28T23:16:41-03:00
 x^{2} + x = 30
 x^{2} + x - 30 = 0
a= 1 b=1 c= -30
Δ = 121
x = -1 + ou - 11/2     x1 = 10/ 2 = 5   x2= -12/ 2 = -6

5 4 5
2014-07-28T23:17:10-03:00
x^2+x=30

x^2+x-30=0

x=\frac{-1+-\sqrt{1^2-4*1*(-30)}}{2*1}=\frac{-1+-\sqrt{1+120}}{2}=\frac{-1+-\sqrt{121}}{2}= \frac{-1+-11}{2}

x'= \frac{-1+11}{2}= \frac{10}{2}=5

x''= \frac{-1-11}{2} = \frac{-12}{2} =-6

S={x ∈ lR | x=-6 ∨ x=5}

Esse número pode ser -6 ou 5.
3 4 3