0=n2+n-132

Solution for 0=n2+n-132 equation:

0=n2+n-132

We move all terms to the left:

0-(n2+n-132)=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-n^2-n+132=0

We add all the numbers together, and all the variables

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