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(5/9)(f-3)=25
We move all terms to the left:
(5/9)(f-3)-(25)=0
Domain of the equation: 9)(f-3)!=0We add all the numbers together, and all the variables
f∈R
(+5/9)(f-3)-25=0
We multiply parentheses ..
(+5f^2+5/9*-3)-25=0
We multiply all the terms by the denominator
(+5f^2+5-25*9*-3)=0
We get rid of parentheses
5f^2+5-3-25*9*=0
We add all the numbers together, and all the variables
5f^2=0
a = 5; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·5·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$f=\frac{-b}{2a}=\frac{0}{10}=0$
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