(n-2)(n+4)=7

Solution for (n-2)(n+4)=7 equation:

(n-2)(n+4)=7

We move all terms to the left:

(n-2)(n+4)-(7)=0

We multiply parentheses ..

(+n^2+4n-2n-8)-7=0

We get rid of parentheses

n^2+4n-2n-8-7=0

We add all the numbers together, and all the variables

n^2+2n-15=0

a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2}
=3
$