(n+2)(n+4)=27

Solution for (n+2)(n+4)=27 equation:

(n+2)(n+4)=27

We move all terms to the left:

(n+2)(n+4)-(27)=0

We multiply parentheses ..

(+n^2+4n+2n+8)-27=0

We get rid of parentheses

n^2+4n+2n+8-27=0

We add all the numbers together, and all the variables

n^2+6n-19=0

a = 1; b = 6; c = -19;
Δ = b2-4ac
Δ = 62-4·1·(-19)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{7}}{2*1}=\frac{-6-4\sqrt{7}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{7}}{2*1}=\frac{-6+4\sqrt{7}}{2}
$