(n-2)(n+2)=27

Solution for (n-2)(n+2)=27 equation:

(n-2)(n+2)=27

We move all terms to the left:

(n-2)(n+2)-(27)=0

We use the square of the difference formula

n^2-4-27=0

We add all the numbers together, and all the variables

n^2-31=0

a = 1; b = 0; c = -31;
Δ = b2-4ac
Δ = 02-4·1·(-31)
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{31}}{2*1}=\frac{0-2\sqrt{31}}{2}
=-\frac{2\sqrt{31}}{2}
=-\sqrt{31}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{31}}{2*1}=\frac{0+2\sqrt{31}}{2}
=\frac{2\sqrt{31}}{2}
=\sqrt{31}
$