Solution:
Given the equation;
[tex]x^2+2x-1=0[/tex]
Using completing the square method; take one to the other side of the equation and change the sign.
[tex]x^2+2x=1[/tex]
Add the square of half of the coefficient of x to both sides of the equation;
[tex]\begin{gathered} x^2+2x+(\frac{1}{2}(2))^2=1+(\frac{1}{2}(2))^2 \\ \\ x^2+2x+1=1+1 \end{gathered}[/tex]
Factorize the left side and simplify the right side;
[tex](x+1)^2=2[/tex]
Thus, the form is;
[tex](x+1)^{2}=2[/tex]
And the solution is;
[tex]\begin{gathered} x=-1\pm\sqrt{2} \\ \\ x=-1+\sqrt{2},x=-1-\sqrt{2} \\ \\ x=0.41,x=-2.41 \end{gathered}[/tex]
