(n-4)(n+4)=27

Solution for (n-4)(n+4)=27 equation:

(n-4)(n+4)=27

We move all terms to the left:

(n-4)(n+4)-(27)=0

We use the square of the difference formula

n^2-16-27=0

We add all the numbers together, and all the variables

n^2-43=0

a = 1; b = 0; c = -43;
Δ = b2-4ac
Δ = 02-4·1·(-43)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{43}}{2*1}=\frac{0-2\sqrt{43}}{2}
=-\frac{2\sqrt{43}}{2}
=-\sqrt{43}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{43}}{2*1}=\frac{0+2\sqrt{43}}{2}
=\frac{2\sqrt{43}}{2}
=\sqrt{43}
$