(b+2)(b+2)=(b-2)(b+2)-32

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Solution for (b+2)(b+2)=(b-2)(b+2)-32 equation: 



(b+2)(b+2)=(b-2)(b+2)-32

We move all terms to the left:

(b+2)(b+2)-((b-2)(b+2)-32)=0

We use the square of the difference formula

b^2+(b+2)(b+2)+4=0

We multiply parentheses ..

b^2+(+b^2+2b+2b+4)+4=0

We get rid of parentheses

b^2+b^2+2b+2b+4+4=0

We add all the numbers together, and all the variables

2b^2+4b+8=0

a = 2; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·2·8
Δ = -48
Delta is less than zero, so there is no solution for the equation

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