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(F)=4F+2/5F
We move all terms to the left:
(F)-(4F+2/5F)=0
Domain of the equation: 5F)!=0We add all the numbers together, and all the variables
F!=0/1
F!=0
F∈R
F-(+4F+2/5F)=0
We get rid of parentheses
F-4F-2/5F=0
We multiply all the terms by the denominator
F*5F-4F*5F-2=0
Wy multiply elements
5F^2-20F^2-2=0
We add all the numbers together, and all the variables
-15F^2-2=0
a = -15; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·(-15)·(-2)
Δ = -120
Delta is less than zero, so there is no solution for the equation
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