F(x)=(x^2+2x+5)/3x

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Solution for F(x)=(x^2+2x+5)/3x equation: 



(F)=(F^2+2F+5)/3F

We move all terms to the left:

(F)-((F^2+2F+5)/3F)=0



Domain of the equation: 3F)!=0

F!=0/1

F!=0

F∈R



We multiply all the terms by the denominator


F*3F)-((F^2+2F+5)=0

Wy multiply elements

3F^2=0

a = 3; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·3·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$F=\frac{-b}{2a}=\frac{0}{6}=0$

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