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

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

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

We multiply parentheses

-2x^2+(x+1)(x+1)+4x=0

We multiply parentheses ..

-2x^2+(+x^2+x+x+1)+4x=0

We get rid of parentheses

-2x^2+x^2+x+x+4x+1=0

We add all the numbers together, and all the variables

-1x^2+6x+1=0

a = -1; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·(-1)·1
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{10}}{2*-1}=\frac{-6-2\sqrt{10}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{10}}{2*-1}=\frac{-6+2\sqrt{10}}{-2}
$