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(5)=3F+9/2F^
We move all terms to the left:
(5)-(3F+9/2F^)=0
Domain of the equation: 2F^)!=0We get rid of parentheses
F!=0/1
F!=0
F∈R
-3F-9/2F^+5=0
We multiply all the terms by the denominator
-3F*2F^+5*2F^-9=0
Wy multiply elements
-6F^2+10F-9=0
a = -6; b = 10; c = -9;
Δ = b2-4ac
Δ = 102-4·(-6)·(-9)
Δ = -116
Delta is less than zero, so there is no solution for the equation
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