0=16t^2+112t+128

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

0=16t^2+112t+128

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -16; b = -112; c = -128;
Δ = b2-4ac
Δ = -1122-4·(-16)·(-128)
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-16\sqrt{17}}{2*-16}=\frac{112-16\sqrt{17}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+16\sqrt{17}}{2*-16}=\frac{112+16\sqrt{17}}{-32}
$