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(F)=6/5F+3
We move all terms to the left:
(F)-(6/5F+3)=0
Domain of the equation: 5F+3)!=0We get rid of parentheses
F∈R
F-6/5F-3=0
We multiply all the terms by the denominator
F*5F-3*5F-6=0
Wy multiply elements
5F^2-15F-6=0
a = 5; b = -15; c = -6;
Δ = b2-4ac
Δ = -152-4·5·(-6)
Δ = 345
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{-(-15)-\sqrt{345}}{2*5}=\frac{15-\sqrt{345}}{10} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{345}}{2*5}=\frac{15+\sqrt{345}}{10} $
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