Question

5(y+3)(y-1)-4=(3+2y)в кв + y в кв

Answer 1

$\displaystyle\bf\\5(y+3)(y-1)-4=(3+2y)^{2} +y^{2} \\\\5\cdot(y^{2} -y+3y-3)-4=9+12y+4y^{2} +y^{2} \\\\5\cdot(y^{2} +2y-3)-4=9+12y+4y^{2} +y^{2} \\\\5y^{2} +10y-15-4=5y^{2} +12y+9\\\\5y^{2} +10y-5y^{2} -12y=9+15+4\\\\-2y=28\\\\\boxed{y=-14}$