(n-11)(n+12)=0

Solution for (n-11)(n+12)=0 equation:

(n-11)(n+12)=0

We multiply parentheses ..

(+n^2+12n-11n-132)=0

We get rid of parentheses

n^2+12n-11n-132=0

We add all the numbers together, and all the variables

n^2+n-132=0

a = 1; b = 1; c = -132;
Δ = b2-4ac
Δ = 12-4·1·(-132)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-23}{2*1}=\frac{-24}{2}
=-12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+23}{2*1}=\frac{22}{2}
=11
$