(4/y+5)-3=(y+6/2y+10)

Solution for (4/y+5)-3=(y+6/2y+10) equation:

D( y )

y = 0

y = 0

y = 0

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

4/y-3+5 = y+(6/2)*y+10 // - y+(6/2)*y+10

4/y-y-((6/2)*y)-10-3+5 = 0

4/y-y-3*y-10-3+5 = 0

4*y^-1-4*y^1-8*y^0 = 0

(4*y^0-4*y^2-8*y^1)/(y^1) = 0 // * y^2

y^1*(4*y^0-4*y^2-8*y^1) = 0

y^1

4-4*y^2-8*y = 0

4-4*y^2-8*y = 0

DELTA = (-8)^2-(-4*4*4)

DELTA = 128

DELTA > 0

y = (128^(1/2)+8)/(-4*2) or y = (8-128^(1/2))/(-4*2)

y = (8*2^(1/2)+8)/(-8) or y = (8-8*2^(1/2))/(-8)

y in { (8*2^(1/2)+8)/(-8), (8-8*2^(1/2))/(-8)} 

y in { (8*2^(1/2)+8)/(-8), (8-8*2^(1/2))/(-8) }