(3/x)+(8/3x)=2

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Solution for (3/x)+(8/3x)=2 equation: 


D( x )

x = 0

x = 0

x = 0

x in (-oo:0) U (0:+oo)

(8/3)*x+3/x = 2 // - 2

(8/3)*x+3/x-2 = 0

8/3*x^1+3*x^-1-2*x^0 = 0

(8/3*x^2-2*x^1+3*x^0)/(x^1) = 0 // * x^2

x^1*(8/3*x^2-2*x^1+3*x^0) = 0

x^1

(8/3)*x^2-2*x+3 = 0

(8/3)*x^2-2*x+3 = 0

DELTA = (-2)^2-(3*4*(8/3))

DELTA = -28

DELTA < 0

x in { } 

x belongs to the empty set

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