The answer is $0$. Write $x^8-x^7+2x^6-2x^5+3x^4-3x^3+4x^2-4x = -5/2$ Then LHS can be factored as $x(x-1)(x^6+2x^4+3x^2+4)$. For $x>1$ or $x<0$, LHS $>0$. Hence no root in $(-\infty, 0) \cup (1, +\infty)$. For $0 \leq x \leq 1$, $x(1-x) \leq 1/4$, with equality at $x=1/2$ and $(x^6+2x^4+3x^2+4) \leq 10$, with equality at $x=1$. Hence, $|x(x-1)(x^6+2x^4+3x^2+4)| < 5/2$ for $x \in [0,1]$. Therefore, no real root in $[0,1]$ either.

Thanks for this good solution.