(y^2-4y+3)/(5y^2-40y+35)

Solution for (y^2-4y+3)/(5y^2-40y+35) equation:

D( y )

5*y^2-(40*y)+35 = 0

5*y^2-(40*y)+35 = 0

5*y^2-(40*y)+35 = 0

5*y^2-40*y+35 = 0

5*y^2-40*y+35 = 0

DELTA = (-40)^2-(4*5*35)

DELTA = 900

DELTA > 0

y = (900^(1/2)+40)/(2*5) or y = (40-900^(1/2))/(2*5)

y = 7 or y = 1

y in (-oo:1) U (1:7) U (7:+oo)

(y^2-(4*y)+3)/(5*y^2-(40*y)+35) = 0

(y^2-4*y+3)/(5*y^2-40*y+35) = 0

y^2-4*y+3 = 0

y^2-4*y+3 = 0

DELTA = (-4)^2-(1*3*4)

DELTA = 4

DELTA > 0

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

y = 3 or y = 1

(y-1)*(y-3) = 0

5*y^2-40*y+35 = 0

5*(y^2-8*y+7) = 0

y^2-8*y+7 = 0

DELTA = (-8)^2-(1*4*7)

DELTA = 36

DELTA > 0

y = (36^(1/2)+8)/(1*2) or y = (8-36^(1/2))/(1*2)

y = 7 or y = 1

5*(y-1)*(y-7) = 0

((y-1)*(y-3))/(5*(y-1)*(y-7)) = 0

( y-1 )

y-1 = 0 // + 1

y = 1

( y-3 )

y-3 = 0 // + 3

y = 3

y in { 1} 

y = 3