108=(2n+4)(2n+1)

Solution for 108=(2n+4)(2n+1) equation:

108=(2n+4)(2n+1)

We move all terms to the left:

108-((2n+4)(2n+1))=0

We multiply parentheses ..

-((+4n^2+2n+8n+4))+108=0


We calculate terms in parentheses: -((+4n^2+2n+8n+4)), so:

(+4n^2+2n+8n+4)

We get rid of parentheses

4n^2+2n+8n+4

We add all the numbers together, and all the variables

4n^2+10n+4

Back to the equation:

-(4n^2+10n+4)


We get rid of parentheses

-4n^2-10n-4+108=0

We add all the numbers together, and all the variables

-4n^2-10n+104=0

a = -4; b = -10; c = +104;
Δ = b2-4ac
Δ = -102-4·(-4)·104
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-42}{2*-4}=\frac{-32}{-8}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+42}{2*-4}=\frac{52}{-8}
=-6+1/2
$