$\displaystyle \log_2x+ \log_2(x+3) = \log_28$
addition of logs gives
$\displaystyle \log_2(x(x+3)) = \log_28$
$\displaystyle \log_2(x^2+3x) = \log_28$
raising both sides to the power of 2
$\displaystyle x^2+3x = 8$
$\displaystyle x^2+3x - 8=0$
now we have a quadratic, do you know how to solve this?