560=16t^2+64t+512

Solution for 560=16t^2+64t+512 equation:

560=16t^2+64t+512

We move all terms to the left:

560-(16t^2+64t+512)=0

We get rid of parentheses

-16t^2-64t-512+560=0

We add all the numbers together, and all the variables

-16t^2-64t+48=0

a = -16; b = -64; c = +48;
Δ = b2-4ac
Δ = -642-4·(-16)·48
Δ = 7168
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{7168}=\sqrt{1024*7}=\sqrt{1024}*\sqrt{7}=32\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-32\sqrt{7}}{2*-16}=\frac{64-32\sqrt{7}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+32\sqrt{7}}{2*-16}=\frac{64+32\sqrt{7}}{-32}
$