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(F)=-2/5F-3
We move all terms to the left:
(F)-(-2/5F-3)=0
Domain of the equation: 5F-3)!=0We get rid of parentheses
F∈R
F+2/5F+3=0
We multiply all the terms by the denominator
F*5F+3*5F+2=0
Wy multiply elements
5F^2+15F+2=0
a = 5; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·5·2
Δ = 185
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{185}}{2*5}=\frac{-15-\sqrt{185}}{10} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{185}}{2*5}=\frac{-15+\sqrt{185}}{10} $
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