[tex]\Large\boxed{\begin{array}{l}\sf\sqrt{3x+6}=x+2\\\rm elevando\,os\,dois\,lados\,ao\,quadrado\,temos:\\\sf(\sqrt{3x+6})^2=(x+2)^2\\\sf x^2+4x+4=3x+6\\\sf x^2+4x-3x+4-6=0\\\sf x^2+x-2=0\\\sf\Delta=b^2-4ac\\\sf\Delta=1^2-4\cdot1\cdot(-2)\\\sf\Delta=1+8\\\sf\Delta=9\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-1\pm\sqrt{9}}{2\cdot1}\\\\\sf x=\dfrac{-1\pm3}{2}\begin{cases}\sf x_1=\dfrac{-1+3}{2}=\dfrac{2}{2}=1\\\\\sf x_2=\dfrac{-1-3}{2}=-\dfrac{4}{2}=-2\end{cases}\end{array}}[/tex]
[tex]\Large\boxed{\begin{array}{l}\underline{\rm Verificac_{\!\!,}\tilde ao:}\\\sf para~x=1:\\\sf \sqrt{3\cdot1+6}=1+2\\\sf \sqrt{3+6}=3\\\sf\sqrt{9}=3\\\sf 3=3\blue{\checkmark}\implies x=1\,\acute e\,ra\acute iz.\\\sf para~x=-2\\\sf\sqrt{3\cdot(-2)+6}=-2+2\\\sf\sqrt{-6+6}=0\\\sf\sqrt{0}=0\\\sf 0=0\blue{\checkmark}\implies x=-2\,\acute e\,ra\acute iz.\\\sf S=\{1,-2\}\end{array}}[/tex]
