(3y-4)(3y-4)=0

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Solution for (3y-4)(3y-4)=0 equation: 



(3y-4)(3y-4)=0

We multiply parentheses ..

(+9y^2-12y-12y+16)=0

We get rid of parentheses

9y^2-12y-12y+16=0

We add all the numbers together, and all the variables

9y^2-24y+16=0

a = 9; b = -24; c = +16;
Δ = b2-4ac
Δ = -242-4·9·16
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{24}{18}=1+1/3$

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