(6/x)+(3/2x)=5/2

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Solution for (6/x)+(3/2x)=5/2 equation: 


D( x )

x = 0

x = 0

x = 0

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

(3/2)*x+6/x = 5/2 // - 5/2

(3/2)*x+6/x-(5/2) = 0

(3/2)*x+6/x-5/2 = 0

3/2*x^1+6*x^-1-5/2*x^0 = 0

(3/2*x^2-5/2*x^1+6*x^0)/(x^1) = 0 // * x^2

x^1*(3/2*x^2-5/2*x^1+6*x^0) = 0

x^1

(3/2)*x^2+(-5/2)*x+6 = 0

(3/2)*x^2+(-5/2)*x+6 = 0

DELTA = (-5/2)^2-(4*6*(3/2))

DELTA = -119/4

DELTA < 0

x in { } 

x belongs to the empty set

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