-128=-16t^2+32t

Solution for -128=-16t^2+32t equation:

-128=-16t^2+32t

We move all terms to the left:

-128-(-16t^2+32t)=0

We get rid of parentheses

16t^2-32t-128=0

a = 16; b = -32; c = -128;
Δ = b2-4ac
Δ = -322-4·16·(-128)
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{-(-32)-96}{2*16}=\frac{-64}{32}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+96}{2*16}=\frac{128}{32}
=4
$