Could you factorize the following? 2(2x-1)^2 - 3(x-3)^2 = 5(x+3)(2x-1)I'm sorry. That was meant to be 3(x+3)^2

2(2x-1)^2 - 3(x-3)^2 =
5(x+3)(2x-1)

We solve the above equation using factors
:

2(2x-1)^2-3(x-3)^2 =
5(x+3)(2x-1)

2(4x^2-4x+1) -3(x^2-6x+9) =
5(2x^2-x+6x-3)

(8-3)x^2+(-8+18)x +2-27 =
10x^2+25x-15

5x^2+10x-25 =
10x^2+25x-15

0 =
(10-5)x^2+(-10+25)x+25-15

5x^2+15x+10 =
0

Divide by
5:

x^2+3x+3

(x+1)(x+2) =
0

x+1 = 0. Or (x+2 = 0.

x=-1.
Or x = -2..

If x-3 is to be replaced by x+3 as suggested in
he next page posted:

Then instead of -3(x-6x+9) , we must
change to  -3(x^2+6x+9).

The correction is  -36x^2 instead
of 26x^2.

This leads to 10x^2+51x +10 =
0

10x^2 +50x+x+10 =
0

x(10x+1)+1(10x+1) =
0

(10x+1)(x+1) = 0

10x+1 = 0.
Or x+1 = 0

10x = -1 or x =
-1/10.

x+1 = 0 gives x =
-1