1/y-3+1/4=-7/2y-6

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Solution for 1/y-3+1/4=-7/2y-6 equation: 


D( y )

y = 0

y = 0

y = 0

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

1/y+1/4-3 = (-7/2)*y-6 // - (-7/2)*y-6

1/y-((-7/2)*y)+1/4-3+6 = 0

(7/2)*y+1/y+1/4-3+6 = 0

7/2*y^1+1*y^-1+13/4*y^0 = 0

(7/2*y^2+13/4*y^1+1*y^0)/(y^1) = 0 // * y^2

y^1*(7/2*y^2+13/4*y^1+1*y^0) = 0

y^1

(7/2)*y^2+(13/4)*y+1 = 0

(7/2)*y^2+(13/4)*y+1 = 0

DELTA = (13/4)^2-(1*4*(7/2))

DELTA = -55/16

DELTA < 0

y in { } 

y belongs to the empty set

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