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What are the solutions to the equation 3y(y)=x^2+12 ?
The solutions to the equation 3y(y)=x^2+12 are x=sqrt(3y(y)-12),x=-sqrt(3y(y)-12)Find the zeros of 3y(y)=x^2+12
The zeros of 3y(y)=x^2+12 are x=sqrt(3y(y)-12),x=-sqrt(3y(y)-12)