Answer:
[tex]\large\boxed{x=-3-\sqrt{15}\ \vee\ x=-3+\sqrt{15}}[/tex]
Step-by-step explanation:
[tex]\text{The quadratic formula:}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\\\\x^2+6x-6=0\\\\a=1,\ b=6,\ c=-6\\\\\text{Substitute:}\\\\x=\dfrac{-6\pm\sqrt{6^2-4(1)(-6)}}{(2)(1)}=\dfrac{-6\pm\sqrt{36+24}}{2}=\dfrac{-6\pm\sqrt{60}}{2}\\\\=\dfrac{-6\pm\sqrt{(4)(15)}}{2}=\dfrac{-6\pm\sqrt4\cdot\sqrt{15}}{2}=\dfrac{-6\pm2\sqrt{15}}{2}\\\\=\dfrac{-6}{2}\pm\dfrac{2\sqrt{15}}{2}=-3\pm\sqrt{15}[/tex]
