Solve for real y y^2+3+ 12/(y^2+3)=7

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We have to solve : y^2 +3 + 12/(y^2+3)=7

y^2 +3 + 12/(y^2+3)=7

=> (y^2 + 3)^2 + 12 = 7( y^2 + 3)

=> y^4 + 9 + 6y^2 + 12 = 7y^2 + 21

=> y^4 - y^2 = 0

=> y^2( y^2 - 1) = 0

y^2 = 0 or y = 0

y^2 - 1 = 0

=> y^2 = 1

=> y = 1 and y = -1

The required real values of y are 0 , -1 and 1

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