Answer:
Step-by-step explanation:
[tex]3x^2 + 12x + 6 = 0\\\\a=3\,,\ b=12\,,\ c=6\\\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{-12\pm\sqrt{12^2-4\cdot3\cdot6}}{2\cdot3}=\dfrac{-12\pm\sqrt{144-72}}{2\cdot3}=\dfrac{-12\pm\sqrt{72}}{2\cdot3}=\dfrac{-12\pm6\sqrt{2}}{6}\\\\x_1=\dfrac{6(-2+\sqrt{2})}{6}=-2+\sqrt2\\\\x_2=\dfrac{6(-2-\sqrt{2})}{6}=-2-\sqrt2[/tex]