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(F)=3/5F+11
We move all terms to the left:
(F)-(3/5F+11)=0
Domain of the equation: 5F+11)!=0We get rid of parentheses
F∈R
F-3/5F-11=0
We multiply all the terms by the denominator
F*5F-11*5F-3=0
Wy multiply elements
5F^2-55F-3=0
a = 5; b = -55; c = -3;
Δ = b2-4ac
Δ = -552-4·5·(-3)
Δ = 3085
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{-(-55)-\sqrt{3085}}{2*5}=\frac{55-\sqrt{3085}}{10} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+\sqrt{3085}}{2*5}=\frac{55+\sqrt{3085}}{10} $
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