(n+2)(n-4)=40

Solution for (n+2)(n-4)=40 equation:

(n+2)(n-4)=40

We move all terms to the left:

(n+2)(n-4)-(40)=0

We multiply parentheses ..

(+n^2-4n+2n-8)-40=0

We get rid of parentheses

n^2-4n+2n-8-40=0

We add all the numbers together, and all the variables

n^2-2n-48=0

a = 1; b = -2; c = -48;
Δ = b2-4ac
Δ = -22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*1}=\frac{-12}{2}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*1}=\frac{16}{2}
=8
$