Question

Решите уравнение
2(2y+3) = y+3+y^2-1

Answer 1

$\displaystyle\bf\\2(2y+3)=y+3+y^{2}-1\\\\4y+6=y+3+y^{2} -1\\\\y^{2}+y-4y-6+3-1=0\\\\y^{2} -3y-4=0\\\\D=(-3)^{2} -4\cdot(-4)=9+16=25=5^{2} \\\\\\y_{1} =\frac{3-5}{2} =-1\\\\\\y_{2} =\frac{3+5}{2} =4\\\\\\Otvet \ : \ -1 \ ; \ 4$