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

Solution for (6-n)(4+n)=0 equation:

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

We add all the numbers together, and all the variables

(-1n+6)(n+4)=0

We multiply parentheses ..

(-1n^2-4n+6n+24)=0

We get rid of parentheses

-1n^2-4n+6n+24=0

We add all the numbers together, and all the variables

-1n^2+2n+24=0

a = -1; b = 2; c = +24;
Δ = b2-4ac
Δ = 22-4·(-1)·24
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-1}=\frac{-12}{-2}
=+6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-1}=\frac{8}{-2}
=-4
$