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4+2p=10((3/5)p-2)
We move all terms to the left:
4+2p-(10((3/5)p-2))=0
Domain of the equation: 5)p-2))!=0We add all the numbers together, and all the variables
p!=0/1
p!=0
p∈R
2p-(10((+3/5)p-2))+4=0
We multiply all the terms by the denominator
2p*5)p-2))-(10((+4*5)p-2))+3=0
We add all the numbers together, and all the variables
2p*5)p-2))-(10(20p-2))+3=0
Wy multiply elements
10p^2=0
a = 10; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·10·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$p=\frac{-b}{2a}=\frac{0}{20}=0$
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