300=x^2+20x

Solution for 300=x^2+20x equation:

300=x^2+20x

We move all terms to the left:

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

We get rid of parentheses

-x^2-20x+300=0

We add all the numbers together, and all the variables

-1x^2-20x+300=0

a = -1; b = -20; c = +300;
Δ = b2-4ac
Δ = -202-4·(-1)·300
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-40}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+40}{2*-1}=\frac{60}{-2}
=-30
$