4(n+2)(n+8)=280

Solution for 4(n+2)(n+8)=280 equation:

4(n+2)(n+8)=280

We move all terms to the left:

4(n+2)(n+8)-(280)=0

We multiply parentheses ..

4(+n^2+8n+2n+16)-280=0

We multiply parentheses

4n^2+32n+8n+64-280=0

We add all the numbers together, and all the variables

4n^2+40n-216=0

a = 4; b = 40; c = -216;
Δ = b2-4ac
Δ = 402-4·4·(-216)
Δ = 5056
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:

$\sqrt{\Delta}=\sqrt{5056}=\sqrt{64*79}=\sqrt{64}*\sqrt{79}=8\sqrt{79}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{79}}{2*4}=\frac{-40-8\sqrt{79}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{79}}{2*4}=\frac{-40+8\sqrt{79}}{8}
$