(5/2z)+(6/z)=1

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Solution for (5/2z)+(6/z)=1 equation: 


D( z )

z = 0

z = 0

z = 0

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

(5/2)*z+6/z = 1 // - 1

(5/2)*z+6/z-1 = 0

5/2*z^1+6*z^-1-1*z^0 = 0

(5/2*z^2-1*z^1+6*z^0)/(z^1) = 0 // * z^2

z^1*(5/2*z^2-1*z^1+6*z^0) = 0

z^1

(5/2)*z^2-z+6 = 0

(5/2)*z^2-z+6 = 0

DELTA = (-1)^2-(4*6*(5/2))

DELTA = -59

DELTA < 0

z in { } 

z belongs to the empty set

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