-16t2+128t-112=0

Solution for -16t2+128t-112=0 equation:

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

a = -16; b = 128; c = -112;
Δ = b2-4ac
Δ = 1282-4·(-16)·(-112)
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-96}{2*-16}=\frac{-224}{-32}
=+7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+96}{2*-16}=\frac{-32}{-32}
=1
$