4n(n+2)=20

Solution for 4n(n+2)=20 equation:

4n(n+2)=20

We move all terms to the left:

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

We multiply parentheses

4n^2+8n-20=0

a = 4; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·4·(-20)
Δ = 384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{6}}{2*4}=\frac{-8-8\sqrt{6}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{6}}{2*4}=\frac{-8+8\sqrt{6}}{8}
$