(7/x)+(10/3x)=(2/9)

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Solution for (7/x)+(10/3x)=(2/9) equation: 


D( x )

x = 0

x = 0

x = 0

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

(10/3)*x+7/x = 2/9 // - 2/9

(10/3)*x+7/x-(2/9) = 0

(10/3)*x+7/x-2/9 = 0

10/3*x^1+7*x^-1-2/9*x^0 = 0

(10/3*x^2-2/9*x^1+7*x^0)/(x^1) = 0 // * x^2

x^1*(10/3*x^2-2/9*x^1+7*x^0) = 0

x^1

(10/3)*x^2+(-2/9)*x+7 = 0

(10/3)*x^2+(-2/9)*x+7 = 0

DELTA = (-2/9)^2-(4*7*(10/3))

DELTA = -7556/81

DELTA < 0

x in { } 

x belongs to the empty set

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