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(F)=3/F-5
We move all terms to the left:
(F)-(3/F-5)=0
Domain of the equation: F-5)!=0We get rid of parentheses
F∈R
F-3/F+5=0
We multiply all the terms by the denominator
F*F+5*F-3=0
We add all the numbers together, and all the variables
5F+F*F-3=0
Wy multiply elements
F^2+5F-3=0
a = 1; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·1·(-3)
Δ = 37
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{-(5)-\sqrt{37}}{2*1}=\frac{-5-\sqrt{37}}{2} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{37}}{2*1}=\frac{-5+\sqrt{37}}{2} $
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