Solve$(log_{2}(x+1))^2=4$ $$(log_{2}(x+1))^2=4$$
$$log_{2}(x+1)*log_{2}(x+1)=log_{2}16$$
$$x^{2}+2x-15=0$$
$$(x+1)*(x+1)=16$$
$$x^{2}+2x+1=16$$
$$x^{2}+2x-15=0$$
$$(x+5)(x-3)=0$$
$$x_1=-5; x_2=3$$
The solution is only $x_1=3$. Is this correct?

$$ (\log_2(x+1))^2 = 4 \Rightarrow \log_2(x+1) = 2 , - 2 $$
$$ \Rightarrow x + 1 = 2^2 , 2^{-2} \Rightarrow x = 3 \space \textrm{or} \space -\frac{3}{4} $$
