(y+2)^2/5=y+2+y^2/3

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Solution for (y+2)^2/5=y+2+y^2/3 equation: 


y in (-oo:+oo)

((y+2)^2)/5 = (y^2)/3+y+2 // - (y^2)/3+y+2

((y+2)^2)/5-((y^2)/3)-y-2 = 0

((y+2)^2)/5+(-y^2)/3-y-2 = 0

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

3*(y+2)^2+5*(-y^2)+3*5*(-y)-2*3*5 = 0

12*y-2*y^2-15*y-30+12 = 0

12-2*y^2-3*y-30 = 0

-2*y^2-3*y-18 = 0

-2*y^2-3*y-18 = 0

-1*(2*y^2+3*y+18) = 0

2*y^2+3*y+18 = 0

DELTA = 3^2-(2*4*18)

DELTA = -135

DELTA < 0

-1 = 0

-1/(3*5) = 0

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

-1 = 0

y belongs to the empty set

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