(2/x)+(3/2x)=-1

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Solution for (2/x)+(3/2x)=-1 equation: 


D( x )

x = 0

x = 0

x = 0

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

(3/2)*x+2/x = -1 // + 1

(3/2)*x+2/x+1 = 0

3/2*x^1+2*x^-1+1*x^0 = 0

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

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

x^1

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

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

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

DELTA = -11

DELTA < 0

x in { } 

x belongs to the empty set

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