(n+12)2=(n+8)(n+18)

Solution for (n+12)2=(n+8)(n+18) equation:

(n+12)2=(n+8)(n+18)

We move all terms to the left:

(n+12)2-((n+8)(n+18))=0

We multiply parentheses

2n-((n+8)(n+18))+24=0

We multiply parentheses ..

-((+n^2+18n+8n+144))+2n+24=0


We calculate terms in parentheses: -((+n^2+18n+8n+144)), so:

(+n^2+18n+8n+144)

We get rid of parentheses

n^2+18n+8n+144

We add all the numbers together, and all the variables

n^2+26n+144

Back to the equation:

-(n^2+26n+144)


We add all the numbers together, and all the variables

2n-(n^2+26n+144)+24=0

We get rid of parentheses

-n^2+2n-26n-144+24=0

We add all the numbers together, and all the variables

-1n^2-24n-120=0

a = -1; b = -24; c = -120;
Δ = b2-4ac
Δ = -242-4·(-1)·(-120)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{6}}{2*-1}=\frac{24-4\sqrt{6}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{6}}{2*-1}=\frac{24+4\sqrt{6}}{-2}
$