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

Solution for 4(n+4)(n-2)=0 equation:

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

We multiply parentheses ..

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

We multiply parentheses

4n^2-8n+16n-32=0

We add all the numbers together, and all the variables

4n^2+8n-32=0

a = 4; b = 8; c = -32;
Δ = b2-4ac
Δ = 82-4·4·(-32)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-24}{2*4}=\frac{-32}{8}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24}{2*4}=\frac{16}{8}
=2
$