20=x^2+4x+4

Solution for 20=x^2+4x+4 equation:

20=x^2+4x+4

We move all terms to the left:

20-(x^2+4x+4)=0

We get rid of parentheses

-x^2-4x-4+20=0

We add all the numbers together, and all the variables

-1x^2-4x+16=0

a = -1; b = -4; c = +16;
Δ = b2-4ac
Δ = -42-4·(-1)·16
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{5}}{2*-1}=\frac{4-4\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{5}}{2*-1}=\frac{4+4\sqrt{5}}{-2}
$