Answer:
[tex]\large\boxed{x=-2-\sqrt2\ or\ x=-2+\sqrt2}[/tex]
Step-by-step explanation:
[tex]3x^2+12x+6=0\qquad\text{divide both sides by 3}\\\\\dfrac{3x^2}{3}+\dfrac{12x}{3}+\dfrac{6}{3}=\dfrac{0}{3}\\\\x^2+4x+2=0\qquad\text{subtract 2 from both sides}\\\\x^2+4x+2-2=0-2\\\\x^2+4x=-2\\\\x^2+2(x)(2)=-2\qquad\text{add}\ 2^2\ \text{to both sides}\\\\x^2+2(x)(2)+2^2=-2+2^2\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+2)^2=-2+4\\\\(x+2)^2=2\iff x+2=\pm\sqrt2\qquad\text{subtract 2 from both sides}\\\\x=-2\pm\sqrt2[/tex]
