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(F)=2F/3F-1
We move all terms to the left:
(F)-(2F/3F-1)=0
Domain of the equation: 3F-1)!=0We get rid of parentheses
F∈R
F-2F/3F+1=0
We multiply all the terms by the denominator
F*3F-2F+1*3F=0
We add all the numbers together, and all the variables
-2F+F*3F+1*3F=0
Wy multiply elements
3F^2-2F+3F=0
We add all the numbers together, and all the variables
3F^2+F=0
a = 3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·3·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*3}=\frac{-2}{6} =-1/3 $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*3}=\frac{0}{6} =0 $
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