(n+3)(n+2)=30

Solution for (n+3)(n+2)=30 equation:

(n+3)(n+2)=30

We move all terms to the left:

(n+3)(n+2)-(30)=0

We multiply parentheses ..

(+n^2+2n+3n+6)-30=0

We get rid of parentheses

n^2+2n+3n+6-30=0

We add all the numbers together, and all the variables

n^2+5n-24=0

a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2}
=-8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2}
=3
$