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(F)=(5F-10/3)
We move all terms to the left:
(F)-((5F-10/3))=0
We add all the numbers together, and all the variables
F-((+5F-10/3))=0
We multiply all the terms by the denominator
F*3))-((+5F-10=0
We add all the numbers together, and all the variables
5F+F*3))-((-10=0
Wy multiply elements
3F^2+5F-10=0
a = 3; b = 5; c = -10;
Δ = b2-4ac
Δ = 52-4·3·(-10)
Δ = 145
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{145}}{2*3}=\frac{-5-\sqrt{145}}{6} $$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{145}}{2*3}=\frac{-5+\sqrt{145}}{6} $
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