-16t^2+140t+56=0

Solution for -16t^2+140t+56=0 equation:

-16t^2+140t+56=0

a = -16; b = 140; c = +56;
Δ = b2-4ac
Δ = 1402-4·(-16)·56
Δ = 23184
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:

$\sqrt{\Delta}=\sqrt{23184}=\sqrt{144*161}=\sqrt{144}*\sqrt{161}=12\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-12\sqrt{161}}{2*-16}=\frac{-140-12\sqrt{161}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+12\sqrt{161}}{2*-16}=\frac{-140+12\sqrt{161}}{-32}
$