#### Answer

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

#### Work Step by Step

Given: $-7x[x(2x-5)-12](9x^2+30x+25)=0$
Need to use factorization to solve the quadratic equation.
This can be factorized as follows:
$-7x(2x+3)(x-4)(3x+5)^2=0$
Now $-7x=0 \implies x=0$
and $(2x+3)=0$
This implies $x=\dfrac{-3}{2}$
and $(x-4)=0 \implies x=4$
and $(3x+5)^2=0$
This implies $x=\dfrac{-5}{3}$
Hence, $x=0,\dfrac{-3}{2},4,\dfrac{-5}{3}$