4(n+5)(n+2)=280^3

Solution for 4(n+5)(n+2)=280^3 equation:

4(n+5)(n+2)=280^3

We move all terms to the left:

4(n+5)(n+2)-(280^3)=0

We add all the numbers together, and all the variables

4(n+5)(n+2)-21952000=0

We multiply parentheses ..

4(+n^2+2n+5n+10)-21952000=0

We multiply parentheses

4n^2+8n+20n+40-21952000=0

We add all the numbers together, and all the variables

4n^2+28n-21951960=0

a = 4; b = 28; c = -21951960;
Δ = b2-4ac
Δ = 282-4·4·(-21951960)
Δ = 351232144
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:

$\sqrt{\Delta}=\sqrt{351232144}=\sqrt{16*21952009}=\sqrt{16}*\sqrt{21952009}=4\sqrt{21952009}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{21952009}}{2*4}=\frac{-28-4\sqrt{21952009}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{21952009}}{2*4}=\frac{-28+4\sqrt{21952009}}{8}
$