F.3 equation

3y² + 6y + 1 = 3(y + m)² + n
3y² + 6y + 1 = 3(y² + 2my + m²) + n
3y² + 6y + 1 = 3y² + 6my + (3m² + n)

Compare the y terms on the both sides :
6y = 6my
m = 1

Compare the constant terms on the both sides :
3m² + n = 1 ...... [1]

Put m = 1 into [1] :
3(1)² + n = 1
n = -2

n=(4ac-b^2)/4a
n=(4*3*1-6*6)/(4*3)
n=-2