(n+2)(n+3)=132

Solution for (n+2)(n+3)=132 equation:

(n+2)(n+3)=132

We move all terms to the left:

(n+2)(n+3)-(132)=0

We multiply parentheses ..

(+n^2+3n+2n+6)-132=0

We get rid of parentheses

n^2+3n+2n+6-132=0

We add all the numbers together, and all the variables

n^2+5n-126=0

a = 1; b = 5; c = -126;
Δ = b2-4ac
Δ = 52-4·1·(-126)
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{-(5)-23}{2*1}=\frac{-28}{2}
=-14
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+23}{2*1}=\frac{18}{2}
=9
$