Решите уравнение

$(t^2+2t)^2-(t+2)(2t^2-t)=6(2t-1)^2\\\\ (t(t+2))^2-t(t+2)(2t-1)-6(2t-1)^2\ |:(2t-1)^2\\\\ \frac{(t(t+2))^2}{(2t-1)^2}- \frac{t(t+2)(2t-1)}{(2t-1)^2}- \frac{6(2t-1)^2}{(2t-1)^2}=0\\\\ ( \frac{(t(t+2))}{(2t-1)})^2- \frac{t(t+2)(2t-1)}{(2t-1)^2}- \frac{6(2t-1)^2}{(2t-1)^2}=0\\\\ \frac{t(t+2)}{2t-1}=y\\\\ y^2-y-6=0\\\\ D=1+24=25; \ \sqrt{D}=5\\\\ y_{1/2}= \frac{1\pm5}{2}\\\\ y_1= \frac{1-5}{2} = -\frac{4}{2}=-2\\\\ y_2= \frac{1+5}{2}= \frac{6}{2}=3$

Обратная замена:

$\frac{t(t+2)}{2t-1}=-2$

ОДЗ:
$2t-1\neq0\\ 2t\neq1\\ t\neq \frac{1}{2}$

$-2(2t-1)=t(t+2)\\\\ -4t+2=t^2+2t\\\\ t^2+2t+4t-2=0\\\\ t^2+6t-2=0\\\\ D=36+8=44; \ \sqrt{D}=2\sqrt{11}\\\\ t_{1/2}= \frac{-6\pm2\sqrt{11}}{2}= \frac{2(-3\pm\sqrt{11})}{2}=-3\pm\sqrt{11}\\\\ t_1=-3-\sqrt{11}\\\\ t_2=\sqrt{11}-3\\\\\\$

$\frac{t(t+2)}{2t-1}=3$

ОДЗ:
$2t-1\neq0\\ 2t\neq1\\ t\neq \frac{1}{2}$

$3(2t-1)=t(t+2)\\\\ 6t-3=t^2+2t\\\\ t^2+2t-6t+3=0\\\\ t^2-4t+3=0\\\\ D=16-12=4; \ \sqrt{D}=2\\\\ t_{1/2}= \frac{4\pm2}{2}=\frac{2(2\pm1)}{2}=2\pm1 \\\\ t_1=2+1=3\\\\ t_1=2-1=1$

Ответ: $\boxed{t_1=-3-\sqrt11; \ t_2=\sqrt{11}+3; \ t_3=1;\ t_4=3}$