0=16t^2-48t-2

Solution for 0=16t^2-48t-2 equation:

0=16t^2-48t-2

We move all terms to the left:

0-(16t^2-48t-2)=0

We add all the numbers together, and all the variables

-(16t^2-48t-2)=0

We get rid of parentheses

-16t^2+48t+2=0

a = -16; b = 48; c = +2;
Δ = b2-4ac
Δ = 482-4·(-16)·2
Δ = 2432
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:

$\sqrt{\Delta}=\sqrt{2432}=\sqrt{64*38}=\sqrt{64}*\sqrt{38}=8\sqrt{38}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{38}}{2*-16}=\frac{-48-8\sqrt{38}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{38}}{2*-16}=\frac{-48+8\sqrt{38}}{-32}
$