-16t^2+128t+112=0

Solution for -16t^2+128t+112=0 equation:

-16t^2+128t+112=0

a = -16; b = 128; c = +112;
Δ = b2-4ac
Δ = 1282-4·(-16)·112
Δ = 23552
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:

$\sqrt{\Delta}=\sqrt{23552}=\sqrt{1024*23}=\sqrt{1024}*\sqrt{23}=32\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-32\sqrt{23}}{2*-16}=\frac{-128-32\sqrt{23}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+32\sqrt{23}}{2*-16}=\frac{-128+32\sqrt{23}}{-32}
$