(b+1)(b-1)=(b+1)(b-1)

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Solution for (b+1)(b-1)=(b+1)(b-1) equation: 



(b+1)(b-1)=(b+1)(b-1)

We move all terms to the left:

(b+1)(b-1)-((b+1)(b-1))=0

We use the square of the difference formula

b^2+b^2-1+1=0

We add all the numbers together, and all the variables

2b^2=0

a = 2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·2·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$b=\frac{-b}{2a}=\frac{0}{4}=0$

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