5/2y+8/y=1

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Solution for 5/2y+8/y=1 equation: 


D( y )

y = 0

y = 0

y = 0

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

(5/2)*y+8/y = 1 // - 1

(5/2)*y+8/y-1 = 0

5/2*y^1+8*y^-1-1*y^0 = 0

(5/2*y^2-1*y^1+8*y^0)/(y^1) = 0 // * y^2

y^1*(5/2*y^2-1*y^1+8*y^0) = 0

y^1

(5/2)*y^2-y+8 = 0

(5/2)*y^2-y+8 = 0

DELTA = (-1)^2-(4*8*(5/2))

DELTA = -79

DELTA < 0

y in { } 

y belongs to the empty set

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