0=-16t^2+96t+408

Solution for 0=-16t^2+96t+408 equation:

0=-16t^2+96t+408

We move all terms to the left:

0-(-16t^2+96t+408)=0

We add all the numbers together, and all the variables

-(-16t^2+96t+408)=0

We get rid of parentheses

16t^2-96t-408=0

a = 16; b = -96; c = -408;
Δ = b2-4ac
Δ = -962-4·16·(-408)
Δ = 35328
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:

$\sqrt{\Delta}=\sqrt{35328}=\sqrt{256*138}=\sqrt{256}*\sqrt{138}=16\sqrt{138}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16\sqrt{138}}{2*16}=\frac{96-16\sqrt{138}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16\sqrt{138}}{2*16}=\frac{96+16\sqrt{138}}{32}
$