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(F)=6F(2/3)^F-5
We move all terms to the left:
(F)-(6F(2/3)^F-5)=0
Domain of the equation: 3)^F-5)!=0We add all the numbers together, and all the variables
F!=0/1
F!=0
F∈R
F-(6F(+2/3)^F-5)=0
We multiply all the terms by the denominator
F*3)^F-5)-(6F(+2=0
Wy multiply elements
3F^2+2=0
a = 3; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·3·2
Δ = -24
Delta is less than zero, so there is no solution for the equation
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