(4y^2+8y+8)^(1/2)+(3y^2+6y+12)^(1/2)=4-2y-y^2 Помогите решить уравнение

$\sqrt{4y^2+8y+8}+ \sqrt{3y^2+6y+12}=4-2y-y^2\\\\ \sqrt{4(y^2+2y+2)}+\sqrt{3(y^2+2y+4)} = -(y^2+2y-4)\\\\ y^2+2y=t\\\\ \sqrt{4(t+2)}+\sqrt{3(t+4)} = 4-t\\\\ 7t+20+2\sqrt{12(t+2)(t+4)}=(4-t)^2\\\\ 2\sqrt{12(t+2)(t+4)}=t^2-15t-4\\\\ 4*12(t+2)(t+4)=(t^2-15t-4)^2\\\\ (23-t)(t-4)^2(t+1)=0\\\\ t=23\\ t=4\\ t=-1\\\\$

$y^2+2y=23\\ y=-2\sqrt{6}-1\\ y=2\sqrt{6}-1\\\\ y^2+2y=4\\ y=-\sqrt{5}-1\\ y=\sqrt{5}-1\\\\ y^2+2y=-1\\ y=-1$
Подходит $y=-1$