505=x^2+20x

Solution for 505=x^2+20x equation:

505=x^2+20x

We move all terms to the left:

505-(x^2+20x)=0

We get rid of parentheses

-x^2-20x+505=0

We add all the numbers together, and all the variables

-1x^2-20x+505=0

a = -1; b = -20; c = +505;
Δ = b2-4ac
Δ = -202-4·(-1)·505
Δ = 2420
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:

$\sqrt{\Delta}=\sqrt{2420}=\sqrt{484*5}=\sqrt{484}*\sqrt{5}=22\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-22\sqrt{5}}{2*-1}=\frac{20-22\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+22\sqrt{5}}{2*-1}=\frac{20+22\sqrt{5}}{-2}
$