The equation [tex]0=-x^2+2x+1[/tex] is equivalent to the equation [tex]x^2-2x-1=0.[/tex]
Solve this equation:
1. [tex]D=b^2-4ac=(-2)^2-4\cdot 1\cdot (-1)=4+4=8.[/tex]
2. [tex]\sqrt{D}=\sqrt{8}=2\sqrt{2}.[/tex]
3.
[tex]x_1=\dfrac{-b+\sqrt{D}}{2a}=\dfrac{2+2\sqrt{2}}{2}=1+\sqrt{2}[/tex]
and
[tex]x_2=\dfrac{-b-\sqrt{D}}{2a}=\dfrac{2-2\sqrt{2}}{2}=1-\sqrt{2}.[/tex]
4. Since [tex]\sqrt{2}>1,[/tex] you have that [tex]1-\sqrt{2}<0.[/tex]
Answer: the positive solution is [tex]x=1+\sqrt{2}.[/tex]
