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(2j+3/3)=(3j-1/6)
We move all terms to the left:
(2j+3/3)-((3j-1/6))=0
We add all the numbers together, and all the variables
(2j+1)-((+3j-1/6))=0
We get rid of parentheses
2j-((+3j-1/6))+1=0
We multiply all the terms by the denominator
2j*6))-((+3j-1+1*6))=0
We add all the numbers together, and all the variables
2j*6))-((3j+5))=0
We add all the numbers together, and all the variables
2j*6))-((3j=0
Wy multiply elements
12j^2=0
a = 12; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·12·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$j=\frac{-b}{2a}=\frac{0}{24}=0$
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