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

Solution for (n-4)(n-3)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

n^2-7n+12=0

a = 1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*1}=\frac{6}{2}
=3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*1}=\frac{8}{2}
=4
$