y^2+4y+4=20

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

y^2+4y+4=20

We move all terms to the left:

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

We add all the numbers together, and all the variables

y^2+4y-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:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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