-16t^2+140t+4=0

Solution for -16t^2+140t+4=0 equation:

-16t^2+140t+4=0

a = -16; b = 140; c = +4;
Δ = b2-4ac
Δ = 1402-4·(-16)·4
Δ = 19856
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:

$\sqrt{\Delta}=\sqrt{19856}=\sqrt{16*1241}=\sqrt{16}*\sqrt{1241}=4\sqrt{1241}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(140)-4\sqrt{1241}}{2*-16}=\frac{-140-4\sqrt{1241}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(140)+4\sqrt{1241}}{2*-16}=\frac{-140+4\sqrt{1241}}{-32}
$