4n^2+4=20

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

4n^2+4=20

We move all terms to the left:

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

We add all the numbers together, and all the variables

4n^2-16=0

a = 4; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·4·(-16)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*4}=\frac{-16}{8}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*4}=\frac{16}{8}
=2
$