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3((-4y-4)/5)+4y=12
We move all terms to the left:
3((-4y-4)/5)+4y-(12)=0
We add all the numbers together, and all the variables
4y+3((-4y-4)/5)-12=0
We multiply all the terms by the denominator
4y*5)+3((-4y-4)-12*5)=0
We add all the numbers together, and all the variables
4y*5)+3((-4y-4)=0
Wy multiply elements
20y^2=0
a = 20; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·20·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$y=\frac{-b}{2a}=\frac{0}{40}=0$
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