(2x+1)(2x+3)(x-1)(x-2) = 150 

FOIL out the first two parentheses on the left

(4x²+8x+3)(x-1)(x-2) = 150

FOIL out the last two parentheses on the left

(4x²+8x+3)(x²-3x+2) = 150

You have to use a generalized version of FOIL", which is:

(F+M+L)(F+M+L) = FF+FM+FL+MF+MM+ML+LF+LM+LL 

Where F means the first term, M means the middle term, 
and L means the last term:

(4x²)(x²)+(4x²)(-3x)+(4x²)(2)+(8x)(x²)+(8x)(-3x)+(8x)(2)+(3)(x²)+(3)(-3x)+(3)(2)

4x4-12x³+8x²+8x³-24x²+16x+3x²-9x+6 = 150

4x4-4x³-13x²+7x-144 = 0

You can factor x² out of the first two terms:

x²[4x²-4x]-13x²+7x-144 = 0

Complete the square in the bracket by adding 1 and subtracting
1 inside the bracket:

x²[4x²-4x+1-1]-13x²+7x-144 = 0

Factor the first three terms inside the bracket as a perfect
square:

x²[(2x-1)²-1]-13x²+7x-144 = 0

Distribute the x² to remove the bracket:

x²(2x-1)²-x²-13x²+7x-144 = 0

Combine the x² terms

x²(2x-1)²-14x²+7x-144 = 0
Factor -7x out of the -14x²+7x

x²(2x-1)²-7x(2x-1)-144 = 0

The first term can be written [x(2x-1)]²

[x(2x-1)]²-7x(2x-1)-144 = 0

Let u = x(2x-1)
   u² = [x(2x-1)]²

u² - 7u - 144 = 0

(u - 16)(u + 9) = 0

u - 16 = 0;  u + 9 = 0
     u = 16;     u = -9

Since u = x(2x-1)

             x(2x-1) = 16;               x(2x-1) = 9
             2x² - x = 16;               2x² - x = 9
             2x²-x-16 = 0;               2x²-x-9 = 0
       x = ;      x = ;
       x = ;                 x = 
       x = ;                   x =  
       x = ;                   x = 

Edwin