(3y+1)(3y+1)=0

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Solution for (3y+1)(3y+1)=0 equation: 



(3y+1)(3y+1)=0

We multiply parentheses ..

(+9y^2+3y+3y+1)=0

We get rid of parentheses

9y^2+3y+3y+1=0

We add all the numbers together, and all the variables

9y^2+6y+1=0

a = 9; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·9·1
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{-6}{18}=-1/3$

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