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

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

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

We add all the numbers together, and all the variables

(-1n+2)(n+4)=0

We multiply parentheses ..

(-1n^2-4n+2n+8)=0

We get rid of parentheses

-1n^2-4n+2n+8=0

We add all the numbers together, and all the variables

-1n^2-2n+8=0

a = -1; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-1)·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-1}=\frac{-4}{-2}
=+2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-1}=\frac{8}{-2}
=-4
$