To solve this equation, we can start by expanding both sides:

(1-3x)(x+1) = (3x-1)(2x+1)
Expanding the left side:
1(x) + 1(-3x) + (-3x)(x) + (-3x)(1)
= x - 3x - 3x^2 - 3x
= -3x^2 - 5x

Expanding the right side:
(3x)(2x) + (3x)(1) + (-1)(2x) + (-1)(1)
= 6x^2 + 3x - 2x - 1
= 6x^2 + x - 1

Now, the equation becomes:
-3x^2 - 5x = 6x^2 + x - 1

Moving all terms to one side of the equation:
-3x^2 - 5x - 6x^2 - x + 1 = 0
-9x^2 - 6x + 1 = 0

Now, we have a quadratic equation. We can solve it by using the quadratic formula:

x = [-(-6) ± sqrt((-6)^2 - 4(-9)(1))] / 2(-9)
x = [6 ± sqrt(36 + 36)] / -18
x = [6 ± sqrt(72)] / -18
x = [6 ± 6√2] / -18
x = (6 + 6√2) / -18 or x = (6 - 6√2) / -18
x = -1 - √2 / 3 or x = -1 + √2 / 3

Therefore, the solutions to the equation (1-3x)(x+1) = (3x-1)(2x+1) are x = -1 - √2 / 3 and x = -1 + √2 / 3.