0=-16t^2+104t+560

Solution for 0=-16t^2+104t+560 equation:

0=-16t^2+104t+560

We move all terms to the left:

0-(-16t^2+104t+560)=0

We add all the numbers together, and all the variables

-(-16t^2+104t+560)=0

We get rid of parentheses

16t^2-104t-560=0

a = 16; b = -104; c = -560;
Δ = b2-4ac
Δ = -1042-4·16·(-560)
Δ = 46656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{46656}=216$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-216}{2*16}=\frac{-112}{32}
=-3+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+216}{2*16}=\frac{320}{32}
=10
$