\[ 4^x - 3(2^{x+3}) + 128 = 0 \] \[ (2^x)^2 - 3(2^x \cdot 2^3) + 128 = 0 \] \[ (2^x)^2 - 3(2^x \cdot 8) + 128 = 0 \] \[ (2^x)^2 - 24(2^x) + 128 = 0 \] Let \( y = 2^x \), then the equation becomes: \[ y^2 - 24y + 128 = 0 \] \[ y^2 - 16y - 8y + 128 = 0 \] \[ y(y - 16) - 8(y - 16) = 0 \] \[ (y - 8)(y - 16) = 0 \] \[ y = 8 \quad \text{or} \quad y = 16 \] If \( y = 8 \), then \( 2^x = 8 = 2^3 \), so \( x = 3 \). If \( y = 16 \), then \( 2^x = 16 = 2^4 \), so \( x = 4 \). The roots are \( x = 3 \) and \( x = 4 \). Sum of roots: \[ 3 + 4 = 7 \]