20=b^2+8b

Solution for 20=b^2+8b equation:

20=b^2+8b

We move all terms to the left:

20-(b^2+8b)=0

We get rid of parentheses

-b^2-8b+20=0

We add all the numbers together, and all the variables

-1b^2-8b+20=0

a = -1; b = -8; c = +20;
Δ = b2-4ac
Δ = -82-4·(-1)·20
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12}{2*-1}=\frac{-4}{-2}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12}{2*-1}=\frac{20}{-2}
=-10
$