480=16d+d^2

Solution for 480=16d+d^2 equation:

480=16d+d^2

We move all terms to the left:

480-(16d+d^2)=0

We get rid of parentheses

-d^2-16d+480=0

We add all the numbers together, and all the variables

-1d^2-16d+480=0

a = -1; b = -16; c = +480;
Δ = b2-4ac
Δ = -162-4·(-1)·480
Δ = 2176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{34}}{2*-1}=\frac{16-8\sqrt{34}}{-2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{34}}{2*-1}=\frac{16+8\sqrt{34}}{-2}
$