(2/y+3)-(3/4-y)=(2y-2/y^2-y-12)

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Solution for (2/y+3)-(3/4-y)=(2y-2/y^2-y-12) equation: 


D( y )

y = 0

y^2 = 0

y = 0

y = 0

y^2 = 0

y^2 = 0

1*y^2 = 0 // : 1

y^2 = 0

y = 0

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

y+2/y-(3/4)+3 = 2*y-y-(2/(y^2))-12 // - 2*y-y-(2/(y^2))-12

y-(2*y)+y+2/y+2/(y^2)-(3/4)+3+12 = 0

y-2*y+y+2/y+2/(y^2)-3/4+3+12 = 0

2*y^-1+2*y^-2+57/4 = 0

t_1 = y^-1

2*t_1^2+2*t_1^1+57/4 = 0

2*t_1^2+2*t_1+57/4 = 0

DELTA = 2^2-(2*4*57/4)

DELTA = -110

DELTA < 0

y belongs to the empty set

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