1/x+4/3x-5/6x+1

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Solution for 1/x+4/3x-5/6x+1 equation: 


D( x )

x = 0

x = 0

x = 0

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

(4/3)*x-((5/6)*x)+1/x+1 = 0

(4/3)*x+(-5/6)*x+1/x+1 = 0

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

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

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

x^1

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

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

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

DELTA = -1

DELTA < 0

x in { } 

x belongs to the empty set

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