0=-16t^2+192t-176

Solution for 0=-16t^2+192t-176 equation:

0=-16t^2+192t-176

We move all terms to the left:

0-(-16t^2+192t-176)=0

We add all the numbers together, and all the variables

-(-16t^2+192t-176)=0

We get rid of parentheses

16t^2-192t+176=0

a = 16; b = -192; c = +176;
Δ = b2-4ac
Δ = -1922-4·16·176
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-192)-160}{2*16}=\frac{32}{32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-192)+160}{2*16}=\frac{352}{32}
=11
$