-16t^2+28t+280=0

Solution for -16t^2+28t+280=0 equation:

-16t^2+28t+280=0

a = -16; b = 28; c = +280;
Δ = b2-4ac
Δ = 282-4·(-16)·280
Δ = 18704
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:

$\sqrt{\Delta}=\sqrt{18704}=\sqrt{16*1169}=\sqrt{16}*\sqrt{1169}=4\sqrt{1169}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{1169}}{2*-16}=\frac{-28-4\sqrt{1169}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{1169}}{2*-16}=\frac{-28+4\sqrt{1169}}{-32}
$