16t^2-192t+112=0

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

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}
$