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(z/5)-6=(2+(1/6)
We move all terms to the left:
(z/5)-6-((2+(1/6))=0
We add all the numbers together, and all the variables
(+z/5)-6-((2+(+1/6))=0
We add all the numbers together, and all the variables
(+z/5)-((2+(+1/6))-6=0
We get rid of parentheses
z/5-((2+(+1/6))-6=0
We calculate fractions
6z^2/()+()/()=0
We add all the numbers together, and all the variables
6z^2/()+1=0
We multiply all the terms by the denominator
6z^2+1*()=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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