0=16t^2+104t+6

Solution for 0=16t^2+104t+6 equation:

0=16t^2+104t+6

We move all terms to the left:

0-(16t^2+104t+6)=0

We add all the numbers together, and all the variables

-(16t^2+104t+6)=0

We get rid of parentheses

-16t^2-104t-6=0

a = -16; b = -104; c = -6;
Δ = b2-4ac
Δ = -1042-4·(-16)·(-6)
Δ = 10432
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:

$\sqrt{\Delta}=\sqrt{10432}=\sqrt{64*163}=\sqrt{64}*\sqrt{163}=8\sqrt{163}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-8\sqrt{163}}{2*-16}=\frac{104-8\sqrt{163}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+8\sqrt{163}}{2*-16}=\frac{104+8\sqrt{163}}{-32}
$