（6x+7）^2(3x+4)(x+1)=6高次方程.求详解以及检验后的最终答案!

（6x+7）^2(3x+4)(x+1)=6
(36x^2+84x+49)(3x^2+7x+4)=6,
令3x^2+7x+4=t,则36x^2+84x+49=12t+1,
原方程成为(12t+1)t=6,
12t^2+t-6=0,
(4t+3)(3t-2)=0,
(1).t=-3/4时,3x^2+7x+4=-3/4,
12x^2+28x+19=0,无实根.
(2).t=2/3时,3x^2+7x+4=2/3,
9x^2+21x+10=0,
(3x+2)(3x+5)=0,
x1=-2/3,x2=-5/3.