Question

решение уравнения 3(3+2 у)^2- 2y(3+2y)+y^2=6

Answer 1

$3(3+2y)^2-2y(3+2y)+y^2=6\ 3(9+12y+4y^2) - 6y-4y^2+y^2=6\ 27+36y+12y^2-6y-3y^2=6\ 9y^2+30y+21=0\ y_{1}=-frac{7}{3}\ y_{2}=-1$