5y+7/2=-3/y-2

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Solution for 5y+7/2=-3/y-2 equation: 


D( y )

y = 0

y = 0

y = 0

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

5*y+7/2 = -3/y-2 // + -3/y-2

5*y-(-3/y)+7/2+2 = 0

5*y+3*y^-1+7/2+2 = 0

5*y^1+3*y^-1+11/2*y^0 = 0

(5*y^2+11/2*y^1+3*y^0)/(y^1) = 0 // * y^2

y^1*(5*y^2+11/2*y^1+3*y^0) = 0

y^1

5*y^2+(11/2)*y+3 = 0

5*y^2+(11/2)*y+3 = 0

DELTA = (11/2)^2-(3*4*5)

DELTA = -119/4

DELTA < 0

y in { } 

y belongs to the empty set

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