2x/x-3+6/x+3=-28/x^2-9

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Solution for 2x/x-3+6/x+3=-28/x^2-9 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+(2*x)/x-3+3 = -28/(x^2)-9 // + -28/(x^2)-9

6/x-(-28/(x^2))+(2*x)/x-3+3+9 = 0

6/x+28*x^-2+(2*x)/x-3+3+9 = 0

6*x^-1+28*x^-2+11 = 0

t_1 = x^-1

28*t_1^2+6*t_1^1+11 = 0

28*t_1^2+6*t_1+11 = 0

DELTA = 6^2-(4*11*28)

DELTA = -1196

DELTA < 0

x belongs to the empty set

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