0=-16t^2+52t+640

Solution for 0=-16t^2+52t+640 equation:

0=-16t^2+52t+640

We move all terms to the left:

0-(-16t^2+52t+640)=0

We add all the numbers together, and all the variables

-(-16t^2+52t+640)=0

We get rid of parentheses

16t^2-52t-640=0

a = 16; b = -52; c = -640;
Δ = b2-4ac
Δ = -522-4·16·(-640)
Δ = 43664
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:

$\sqrt{\Delta}=\sqrt{43664}=\sqrt{16*2729}=\sqrt{16}*\sqrt{2729}=4\sqrt{2729}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4\sqrt{2729}}{2*16}=\frac{52-4\sqrt{2729}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4\sqrt{2729}}{2*16}=\frac{52+4\sqrt{2729}}{32}
$