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Solution

Given: $4^x - 3(2^{x+3}) + 128 = 0$
Writing in terms of $2^x$:
$(2^x)^2 - 3 \cdot 2^x \cdot 2^3 + 128 = 0$
$(2^x)^2 - 24(2^x) + 128 = 0$
Let $t = 2^x$
$\Rightarrow t^2 - 24t + 128 = 0$
$\Rightarrow (t - 8)(t - 16) = 0$
$\Rightarrow t = 8$ or $t = 16$
When $t = 8$:
$2^x = 8 = 2^3 \Rightarrow x = 3$
When $t = 16$:
$2^x = 16 = 2^4 \Rightarrow x = 4$
Sum of roots $= 3 + 4$
$\therefore \boxed{\text{Sum of roots} = 7}$