The sum of all the solutions of log 2(x^2-x)=1 is

The equation gives $x^2-x=2$, i.e. $x^2-x-2=(x-2)(x+1)=0$, so $x=2$ or $x=-1$. For both values $x^2-x=2>0$, so the logarithm is defined and both are valid. Their sum is $1$, which is also read directly as $-(-1)/1$. Rejecting $-1$ as 'negative' is the usual wrong step, since the argument, not $x$, must be positive.