0=16t^2+16t-480

Solution for 0=16t^2+16t-480 equation:

0=16t^2+16t-480

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-16t^2-16t+480=0

a = -16; b = -16; c = +480;
Δ = b2-4ac
Δ = -162-4·(-16)·480
Δ = 30976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{30976}=176$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-176}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+176}{2*-16}=\frac{192}{-32}
=-6
$