To solve this equation, we will first substitute (y = (3-2x)^2).

This gives us:
(y^2 - y - 72 = 0)

Now, we can factor this quadratic equation:
((y - 9)(y + 8) = 0)

Setting each factor equal to zero gives us:
(y - 9 = 0 \Rightarrow y = 9)
(y + 8 = 0 \Rightarrow y = -8)

Now, we substitute back in (y = (3-2x)^2):
((3-2x)^2 = 9) or ((3-2x)^2 = -8)

Solving these equations gives us:
((3-2x) = 3) or ((3-2x) = -3)
(3-2x = 3) or (3-2x = -3)
(-2x = 0) or (-2x = -6)
(x = 0) or (x = 3)

Therefore, the solutions to the equation ((3-2x)^4-(3-2x)^2-72=0) are (x = 0) and (x = 3).