8(n-2)=(n+4)(n-4)

Solution for 8(n-2)=(n+4)(n-4) equation:

8(n-2)=(n+4)(n-4)

We move all terms to the left:

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

We use the square of the difference formula

n^2+8(n-2)+16=0

We multiply parentheses

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

We add all the numbers together, and all the variables

n^2+8n=0

a = 1; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*1}=\frac{-16}{2}
=-8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*1}=\frac{0}{2}
=0
$