16=4(n+2)n=

Solution for 16=4(n+2)n= equation:

16=4(n+2)n=

We move all terms to the left:

16-(4(n+2)n)=0


We calculate terms in parentheses: -(4(n+2)n), so:

4(n+2)n

We multiply parentheses

4n^2+8n

Back to the equation:

-(4n^2+8n)


We get rid of parentheses

-4n^2-8n+16=0

a = -4; b = -8; c = +16;
Δ = b2-4ac
Δ = -82-4·(-4)·16
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{5}}{2*-4}=\frac{8-8\sqrt{5}}{-8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{5}}{2*-4}=\frac{8+8\sqrt{5}}{-8}
$