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z/3=(5/6)z+25
We move all terms to the left:
z/3-((5/6)z+25)=0
Domain of the equation: 6)z+25)!=0We add all the numbers together, and all the variables
z!=0/1
z!=0
z∈R
z/3-((+5/6)z+25)=0
We calculate fractions
6z^2/18z+()/18z=0
We multiply all the terms by the denominator
6z^2+()=0
We add all the numbers together, and all the variables
6z^2=0
a = 6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·6·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$z=\frac{-b}{2a}=\frac{0}{12}=0$
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