(n-5)(n-7)=0

Solution for (n-5)(n-7)=0 equation:

(n-5)(n-7)=0

We multiply parentheses ..

(+n^2-7n-5n+35)=0

We get rid of parentheses

n^2-7n-5n+35=0

We add all the numbers together, and all the variables

n^2-12n+35=0

a = 1; b = -12; c = +35;
Δ = b2-4ac
Δ = -122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2}{2*1}=\frac{10}{2}
=5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2}{2*1}=\frac{14}{2}
=7
$