(n-5)(n-0)=2

Solution for (n-5)(n-0)=2 equation:

(n-5)(n-0)=2

We move all terms to the left:

(n-5)(n-0)-(2)=0

We add all the numbers together, and all the variables

(n-5)(+n)-2=0

We multiply parentheses ..

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

We get rid of parentheses

n^2-5n-2=0

a = 1; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·1·(-2)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{33}}{2*1}=\frac{5-\sqrt{33}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{33}}{2*1}=\frac{5+\sqrt{33}}{2}
$