I need help solve for x x(x-4)(3x+2)=240

x(x-4)(3x+2) = 240

First we
will open brackets:

(x^2 - 4x)(3x+2) =
240

==> 3x^3 + 2x^2 -12x^2 - 8x - 240 =
0

==> 3x^3 -10x^2 - 8x - 240
=0

Now we will try and substitute with 240
factors.

==> x= 1 ==> 3-10-8 -240 =
-255

==> x= 3 ==> 27 - 90 - 8 - 240 =
-311

==> x= 6 ==> 648 - 360 - 48 - 240 =
0

Then we find that x= 6 is one of the roots ... Then (x-6)
is a facto.

Now we will divide the equation by (x-6) to
find the other factors.

==> (x-6)(3x^2 +8x + 40) =
0

==> x1= 6

Now we will
use the quadratic equation to find the other 2
roots.

==> x2= ( -8 + sqrt( 64-4*3*40)/ 2*3 = ( -8 +
sqrt(416)*i)/6= -8+4sqrt26 )/ 6 = (4/3) +
(2sqrt26/3)*i

==> x2= (4/3) -
(2sqrt26/3)*i

Then the roots are: x= { 6,
(-4/3)+(2sqrt26/3)*i  , (-4/3 - (sqrt26/3)*i}