(n-3)(n-4)=20

Solution for (n-3)(n-4)=20 equation:

(n-3)(n-4)=20

We move all terms to the left:

(n-3)(n-4)-(20)=0

We multiply parentheses ..

(+n^2-4n-3n+12)-20=0

We get rid of parentheses

n^2-4n-3n+12-20=0

We add all the numbers together, and all the variables

n^2-7n-8=0

a = 1; b = -7; c = -8;
Δ = b2-4ac
Δ = -72-4·1·(-8)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*1}=\frac{-2}{2}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*1}=\frac{16}{2}
=8
$