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(F)=-5/3F-8
We move all terms to the left:
(F)-(-5/3F-8)=0
Domain of the equation: 3F-8)!=0We get rid of parentheses
F∈R
F+5/3F+8=0
We multiply all the terms by the denominator
F*3F+8*3F+5=0
Wy multiply elements
3F^2+24F+5=0
a = 3; b = 24; c = +5;
Δ = b2-4ac
Δ = 242-4·3·5
Δ = 516
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}$$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{129}}{2*3}=\frac{-24-2\sqrt{129}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{129}}{2*3}=\frac{-24+2\sqrt{129}}{6} $
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