0=16t^2+128t-192

Solution for 0=16t^2+128t-192 equation:

0=16t^2+128t-192

We move all terms to the left:

0-(16t^2+128t-192)=0

We add all the numbers together, and all the variables

-(16t^2+128t-192)=0

We get rid of parentheses

-16t^2-128t+192=0

a = -16; b = -128; c = +192;
Δ = b2-4ac
Δ = -1282-4·(-16)·192
Δ = 28672
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:

$\sqrt{\Delta}=\sqrt{28672}=\sqrt{4096*7}=\sqrt{4096}*\sqrt{7}=64\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-64\sqrt{7}}{2*-16}=\frac{128-64\sqrt{7}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+64\sqrt{7}}{2*-16}=\frac{128+64\sqrt{7}}{-32}
$