0=16t^2+192t+112

Solution for 0=16t^2+192t+112 equation:

0=16t^2+192t+112

We move all terms to the left:

0-(16t^2+192t+112)=0

We add all the numbers together, and all the variables

-(16t^2+192t+112)=0

We get rid of parentheses

-16t^2-192t-112=0

a = -16; b = -192; c = -112;
Δ = b2-4ac
Δ = -1922-4·(-16)·(-112)
Δ = 29696
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:

$\sqrt{\Delta}=\sqrt{29696}=\sqrt{1024*29}=\sqrt{1024}*\sqrt{29}=32\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-192)-32\sqrt{29}}{2*-16}=\frac{192-32\sqrt{29}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-192)+32\sqrt{29}}{2*-16}=\frac{192+32\sqrt{29}}{-32}
$