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(F)=2/3F+5(F)=-1/3F+1
We move all terms to the left:
(F)-(2/3F+5(F))=0
Domain of the equation: 3F+5F)!=0We add all the numbers together, and all the variables
F∈R
F-(+5F+2/3F)=0
We get rid of parentheses
F-5F-2/3F=0
We multiply all the terms by the denominator
F*3F-5F*3F-2=0
Wy multiply elements
3F^2-15F^2-2=0
We add all the numbers together, and all the variables
-12F^2-2=0
a = -12; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·(-12)·(-2)
Δ = -96
Delta is less than zero, so there is no solution for the equation
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