(7/x)+(5/6x-4)=(4/6x-4)

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Solution for (7/x)+(5/6x-4)=(4/6x-4) equation: 


D( x )

x = 0

x = 0

x = 0

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

(5/6)*x+7/x-4 = (4/6)*x-4 // - (4/6)*x-4

(5/6)*x-((4/6)*x)+7/x-4+4 = 0

(5/6)*x+(-2/3)*x+7/x-4+4 = 0

1/6*x^1+7*x^-1 = 0

(1/6*x^2+7*x^0)/(x^1) = 0 // * x^2

x^1*(1/6*x^2+7*x^0) = 0

x^1

(1/6)*x^2+7 = 0

(1/6)*x^2+7 = 0

DELTA = 0^2-(4*7*(1/6))

DELTA = -14/3

DELTA < 0

x in { } 

x belongs to the empty set

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