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(F)=5/3F-1
We move all terms to the left:
(F)-(5/3F-1)=0
Domain of the equation: 3F-1)!=0We get rid of parentheses
F∈R
F-5/3F+1=0
We multiply all the terms by the denominator
F*3F+1*3F-5=0
Wy multiply elements
3F^2+3F-5=0
a = 3; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·3·(-5)
Δ = 69
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{69}}{2*3}=\frac{-3-\sqrt{69}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{69}}{2*3}=\frac{-3+\sqrt{69}}{6} $
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