(20+y)y=-20

Solution for (20+y)y=-20 equation:

(20+y)y=-20

We move all terms to the left:

(20+y)y-(-20)=0

We add all the numbers together, and all the variables

(y+20)y-(-20)=0

We add all the numbers together, and all the variables

(y+20)y+20=0

We multiply parentheses

y^2+20y+20=0

a = 1; b = 20; c = +20;
Δ = b2-4ac
Δ = 202-4·1·20
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{5}}{2*1}=\frac{-20-8\sqrt{5}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{5}}{2*1}=\frac{-20+8\sqrt{5}}{2}
$