4/x-5/x^2=15

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Solution for 4/x-5/x^2=15 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)

4/x-(5/(x^2)) = 15 // - 15

4/x-(5/(x^2))-15 = 0

4/x-5*x^-2-15 = 0

4*x^-1-5*x^-2-15 = 0

t_1 = x^-1

4*t_1^1-5*t_1^2-15 = 0

4*t_1-5*t_1^2-15 = 0

DELTA = 4^2-(-15*(-5)*4)

DELTA = -284

DELTA < 0

x belongs to the empty set

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