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1=6 math problem
∵(6x+7)^2(3x+4)(x+ 1)=6,

∴(36x^2+84x+49)(3x^2+7x+4)=6,

Let y = 3x 2+7x+4, then

( 12y+ 1)y=6,

12y^2+y-6=0,

Y=2/3 or -3/4,

That is 3x 2+7x+4 = 2/3, or 3x 2+7x+4 =-3/4,

Arrange 3x 2+7x+ 10/3 = 0 or 3x 2+7x+ 19/4 = 0,

The solution is x 1 =-2/3, x2 =-5/3, x3 = (-7+2 √ 2i)/6, x4 = (-7-2 √ 2i)/6.