(7/2x-5)+(2/x+3)

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Solution for (7/2x-5)+(2/x+3) equation: 


D( x )

x = 0

x = 0

x = 0

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

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

7/2*x^1+2*x^-1-2*x^0 = 0

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

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

x^1

(7/2)*x^2-2*x+2 = 0

(7/2)*x^2-2*x+2 = 0

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

DELTA = -24

DELTA < 0

x in { } 

x belongs to the empty set

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