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

Solution for (2+1n)(4-1n)=0 equation:

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

We add all the numbers together, and all the variables

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