(n-5)(5+n)=0

Solution for (n-5)(5+n)=0 equation:

(n-5)(5+n)=0

We add all the numbers together, and all the variables

(n-5)(n+5)=0

We use the square of the difference formula

n^2-25=0

a = 1; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·1·(-25)
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{-(0)-10}{2*1}=\frac{-10}{2}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*1}=\frac{10}{2}
=5
$