(5y/y-2)-(10/y+2)=(40/y^2-4)

Solution for (5y/y-2)-(10/y+2)=(40/y^2-4) 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)

(5*y)/y-(10/y)-2-2 = 40/(y^2)-4 // - 40/(y^2)-4

(5*y)/y-(10/y)-(40/(y^2))-2-2+4 = 0

(5*y)/y-10*y^-1-40*y^-2-2-2+4 = 0

5-10*y^-1-40*y^-2 = 0

t_1 = y^-1

5-40*t_1^2-10*t_1^1 = 0

5-40*t_1^2-10*t_1 = 0

DELTA = (-10)^2-(-40*4*5)

DELTA = 900

DELTA > 0

t_1 = (900^(1/2)+10)/(-40*2) or t_1 = (10-900^(1/2))/(-40*2)

t_1 = -1/2 or t_1 = 1/4

t_1 = -1/2

y^-1+1/2 = 0

1*y^-1 = -1/2 // : 1

y^-1 = -1/2

-1 < 0

1/(y^1) = -1/2 // * y^1

1 = -1/2*y^1 // : -1/2

-2 = y^1

y = -2

t_1 = 1/4

y^-1-1/4 = 0

1*y^-1 = 1/4 // : 1

y^-1 = 1/4

-1 < 0

1/(y^1) = 1/4 // * y^1

1 = 1/4*y^1 // : 1/4

4 = y^1

y = 4

y in { -2, 4 }