[tex]\it x^2+4x+4=x^2+2\cdot2\cdot x+2^2=(x+2)^2\\ \\ 19+8\sqrt3=16+3+8\sqrt3=4^2+2\cdot4\cdot \sqrt3+(\sqrt3)^2=(4+\sqrt3)^2[/tex]
Acum, ecuația devine:
[tex]\it (x+2)^2+(4+\sqrt3)^2 \Rightarrow \sqrt{(x+2)^2}=\sqrt{(4+\sqrt3)^2} \Rightarrow |x+2|=4+\sqrt3 \Rightarrow \\ \\ \Rightarrow x+2=\pm(4+\sqrt2) \Rightarrow \begin{cases}\it x+2=-4-\sqrt3|_{-2} \Rightarrow x_1=-6-\sqrt3\\ \\ \it x+2=4+\sqrt3|_{-2} \Rightarrow x_2=2+\sqrt3\end{cases}[/tex]
