(4x+4)(x+1)=0

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Solution for (4x+4)(x+1)=0 equation: 



(4x+4)(x+1)=0

We multiply parentheses ..

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

We get rid of parentheses

4x^2+4x+4x+4=0

We add all the numbers together, and all the variables

4x^2+8x+4=0

a = 4; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·4·4
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-8}{8}=-1$

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