(4/x+5)+(7/3x-4)

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Solution for (4/x+5)+(7/3x-4) equation: 


D( x )

x = 0

x = 0

x = 0

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

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

7/3*x^1+4*x^-1+1*x^0 = 0

(7/3*x^2+1*x^1+4*x^0)/(x^1) = 0 // * x^2

x^1*(7/3*x^2+1*x^1+4*x^0) = 0

x^1

(7/3)*x^2+x+4 = 0

(7/3)*x^2+x+4 = 0

DELTA = 1^2-(4*4*(7/3))

DELTA = -109/3

DELTA < 0

x in { } 

x belongs to the empty set

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