441=x^2+2x

Solution for 441=x^2+2x equation:

441=x^2+2x

We move all terms to the left:

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

We get rid of parentheses

-x^2-2x+441=0

We add all the numbers together, and all the variables

-1x^2-2x+441=0

a = -1; b = -2; c = +441;
Δ = b2-4ac
Δ = -22-4·(-1)·441
Δ = 1768
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:

$\sqrt{\Delta}=\sqrt{1768}=\sqrt{4*442}=\sqrt{4}*\sqrt{442}=2\sqrt{442}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{442}}{2*-1}=\frac{2-2\sqrt{442}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{442}}{2*-1}=\frac{2+2\sqrt{442}}{-2}
$