Let u = x - 1/2

(u + 1/2)^6 = (u - 1/2)^6

Multiply by 2^6 to avoid typing a lot of fractions.

(2u + 1)^6 = (2u - 1)^6

The even powered terms will cancel out.  Then the RHS will be all negative terms, exactly the opposite of the LHS.  Moving the RHS to the LHS doubles the LHS and sets it equal to zero.  Then divide LHS by 2.  Or just compute LHS and set it to zero.

6(2u)^5 + 20(2u)^3 + 6(2u) = 0

192 u^5 + 160 u^3 + 12 u = 0

u(48 u^4 + 40 u^2 + 3) = 0   quadratic in u^2

u = 0 (x = 1/2) is one solution

u^2 = ( -40 ± sqrt(1600 - 576) )/96

u^2 = ( -40 ± 32 )/96 = {- 72/96, - 8/96}

u^2 = {- 3/4, - 1/12}

u = {√3/2 i, -√3/2 i, √3/6 i, √3/6 i}

x = {1/2, 1/2 + √3/2 i, 1/2 - √3/2 i, 1/2 + √3/6 i, 1/2 - √3/6 i}