Решите уравнение 4*(3y+1)^2-27=(4y+9)(4y-9)+2*(5y+2)(2y-7).

\[4 \cdot (3y + 1)^{2} - 27 =\]

\[= (4y + 9)(4y - 9) + 2 \cdot (5y + 2)(2y - 7)\]

\[4 \cdot \left( 9y^{2} + 6y + 1 \right) - 27 =\]

\[= 16y^{2} - 81 + 2 \cdot \left( 10y^{2} - 31y - 14 \right)\]

\[36y^{2} + 24y + 4 - 27 =\]

\[= 16y^{2} - 81 + 20y^{2} - 62y - 28\]

\[86y = - 86\]

\[y = - 1\]

\[Ответ:\ y = - 1.\ \ \]