-40/y^2-12y+27=2y/y-3

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Solution for -40/y^2-12y+27=2y/y-3 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)

27-(12*y)-40/(y^2) = (2*y)/y-3 // - (2*y)/y-3

3-(12*y)-40/(y^2)-((2*y)/y)+27 = 0

3-12*y-40/(y^2)-2+27 = 0

28*y^0-12*y^1-40*y^-2 = 0

(28*y^2-12*y^3-40*y^0)/(y^2) = 0 // * y^4

y^2*(28*y^2-12*y^3-40*y^0) = 0

y^2

28*y^2-12*y^3-40 = 0

{ 1, -1, 2, -2, 4, -4, 5, -5, 8, -8, 10, -10, 20, -20, 40, -40 }

1

y = 1

28*y^2-12*y^3-40 = -24

1

-1

y = -1

28*y^2-12*y^3-40 = 0

-1

y+1

40*y-12*y^2-40

28*y^2-12*y^3-40

y+1

12*y^3+12*y^2

40*y^2-40

-40*y^2-40*y

-40*y-40

40*y+40

0

40*y-12*y^2-40 = 0

DELTA = 40^2-(-40*(-12)*4)

DELTA = -320

DELTA < 0

y in { -1}

y = -1

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