z/3=(5/6)z+25

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Solution for z/3=(5/6)z+25 equation: 



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)!=0

z!=0/1

z!=0

z∈R



We add all the numbers together, and all the variables

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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