0=-16t^2+96t+48

Solution for 0=-16t^2+96t+48 equation:

0=-16t^2+96t+48

We move all terms to the left:

0-(-16t^2+96t+48)=0

We add all the numbers together, and all the variables

-(-16t^2+96t+48)=0

We get rid of parentheses

16t^2-96t-48=0

a = 16; b = -96; c = -48;
Δ = b2-4ac
Δ = -962-4·16·(-48)
Δ = 12288
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:

$\sqrt{\Delta}=\sqrt{12288}=\sqrt{4096*3}=\sqrt{4096}*\sqrt{3}=64\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-64\sqrt{3}}{2*16}=\frac{96-64\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+64\sqrt{3}}{2*16}=\frac{96+64\sqrt{3}}{32}
$