3x=1+(-8/x)

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Solution for 3x=1+(-8/x) equation: 


D( x )

x = 0

x = 0

x = 0

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

3*x = 1-8/x // - 1-8/x

3*x-(-8/x)-1 = 0

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

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

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

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

x^1

3*x^2-x+8 = 0

3*x^2-x+8 = 0

DELTA = (-1)^2-(3*4*8)

DELTA = -95

DELTA < 0

x in { } 

x belongs to the empty set

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