(9/7z)+(3/z)=1

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Solution for (9/7z)+(3/z)=1 equation: 


D( z )

z = 0

z = 0

z = 0

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

(9/7)*z+3/z = 1 // - 1

(9/7)*z+3/z-1 = 0

9/7*z^1+3*z^-1-1*z^0 = 0

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

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

z^1

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

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

DELTA = (-1)^2-(3*4*(9/7))

DELTA = -101/7

DELTA < 0

z in { } 

z belongs to the empty set

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