(8/x)+(5/5x)=(6/x-2)

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Solution for (8/x)+(5/5x)=(6/x-2) equation: 


D( x )

x = 0

x = 0

x = 0

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

(5/5)*x+8/x = 6/x-2 // - 6/x-2

(5/5)*x+8/x-(6/x)+2 = 0

(5/5)*x+8/x-6*x^-1+2 = 0

1*x^1+2*x^-1+2*x^0 = 0

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

x^1*(1*x^2+2*x^1+2*x^0) = 0

x^1

x^2+2*x+2 = 0

x^2+2*x+2 = 0

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

DELTA = -4

DELTA < 0

x in { } 

x belongs to the empty set

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