срочно,нужно решить систему(2x+4)^2=3y(4x+2)^2=3y

Решите задачу:

$\left \{ {{(2x+4)^2=3y,} \atop {(4x+2)^2=3y;}} \right. \left \{ {{(2x+4)^2=3y,} \atop {(4x+2)^2=(2x+4)^2;}} \right. \left \{ {{(2x+4)^2=3y,} \atop { \left [ {{4x+2=2x+4,} \atop {4x+2=-(2x+4);}} \right.}} \right. \left [ {{ \left \{ {{(2x+4)^2=3y,} \atop {2x=2;}} \right. } \atop { \left \{ {{(2x+4)^2=3y,} \atop {6x=-6;}} \right.}} \right. \\$
$\left [ {{ \left \{ {{x=1,} \atop {3y=36;}} \right. } \atop { \left \{ {{x=-1,} \atop {3y=4;}} \right.}} \right. \left [ {{ \left \{ {{x=1,} \atop {y=12;}} \right. } \atop { \left \{ {{x=-1,} \atop {y=1 \frac{1}{3} ;}} \right.}} \right. \\ (1;12), (-1;1 \frac{1}{3}).$