[tex]METHOD\ 1:\\\\Use:\\\\(*)\qquad(a+b)^2=a^2+2ab+b^2\\---------------------\\\\x^2+4x-6=0\qquad\text{add 6 to both sides}\\\\x^2+4x=6\\\\x^2+2(x)(2)=6\qquad\text{add}\ 2^2\ \text{to both sides}\\\\\underbrace{x^2+2(x)(2)+2^2}_{(*)}=6+2^2\\\\(x+2)^2=6+4\\\\(x+2)^2=10\Rightarrow x+2=\pm\sqrt{10}\qquad\text{subtract 2 from both sides}\\\\\boxed{x=-2-\sqrt{10}\ \vee\ x=-2+\sqrt{10}}[/tex]
[tex]METHOD\ 2:\\\\\text{use the quadratic formula:}\\\\ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta>0,\ \text{the equation has two solutions}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}\\\\if\ \Delta=0,\ \text{then the equation has one solution:}\ x=\dfrac{-b}{2a}\\\\if\ \Delta<0,\ \text{then the equation has not solution}[/tex]
[tex]x^2+4x-6=0\\\\a=1,\ b=4,\ c=-6\\\\\Delta=4^2-4(1)(-6)=16+24=40>0\\\\\sqrt\Delta=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}\\\\x_1=\dfrac{-4-2\sqrt{10}}{2(1)}=\dfrac{-4}{2}-\dfrac{2\sqrt{10}}{2}=-2-\sqrt{10}\\\\x_2=\dfrac{-4+2\sqrt{10}}{2(1)}=\dfrac{-4}{2}+\dfrac{2\sqrt{10}}{2}=-2+\sqrt{10}[/tex]
