#### Answer

$x=0,\dfrac{-4}{3},2,\dfrac{4}{5}$

#### Work Step by Step

Given: $-4x[x(3x-2)-8](25x^2-40x+16)=0$
Need to use factorization to solve the quadratic equation.
This can be factorized as follows:
$-4x(3x+4)(x-2)(5x-4)^2=0$
Now $-4x=0 \implies x=0$
and $(3x+4)=0$
This implies $x=\dfrac{-4}{3}$
and $(x-2)=0 \implies x=2$
and $(5x-4)^2=0$
This implies $x=\dfrac{4}{5}$
Hence, $x=0,\dfrac{-4}{3},2,\dfrac{4}{5}$