(6/x-6)+(4/x+3)=(9x^2/x^2-3x-18)

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Solution for (6/x-6)+(4/x+3)=(9x^2/x^2-3x-18) equation: 


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

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

6/x+4/x-6+3 = (9*x^2)/(x^2)-(3*x)-18 // - (9*x^2)/(x^2)-(3*x)-18

3*x+6/x+4/x-((9*x^2)/(x^2))-6+3+18 = 0

3*x+6/x+4/x-9-6+3+18 = 0

3*x^1+10*x^-1+6*x^0 = 0

(3*x^2+6*x^1+10*x^0)/(x^1) = 0 // * x^2

x^1*(3*x^2+6*x^1+10*x^0) = 0

x^1

3*x^2+6*x+10 = 0

3*x^2+6*x+10 = 0

DELTA = 6^2-(3*4*10)

DELTA = -84

DELTA < 0

x in { } 

x belongs to the empty set

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