(n-2)(n-3)=12

Solution for (n-2)(n-3)=12 equation:

(n-2)(n-3)=12

We move all terms to the left:

(n-2)(n-3)-(12)=0

We multiply parentheses ..

(+n^2-3n-2n+6)-12=0

We get rid of parentheses

n^2-3n-2n+6-12=0

We add all the numbers together, and all the variables

n^2-5n-6=0

a = 1; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·1·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*1}=\frac{-2}{2}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*1}=\frac{12}{2}
=6
$