x^2-(2x+1)=0

Solution for x^2-(2x+1)=0 equation:

x^2-(2x+1)=0

We get rid of parentheses

x^2-2x-1=0

a = 1; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·1·(-1)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*1}=\frac{2-2\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*1}=\frac{2+2\sqrt{2}}{2}
$