0=-16t^2+96t+27

Solution for 0=-16t^2+96t+27 equation:

0=-16t^2+96t+27

We move all terms to the left:

0-(-16t^2+96t+27)=0

We add all the numbers together, and all the variables

-(-16t^2+96t+27)=0

We get rid of parentheses

16t^2-96t-27=0

a = 16; b = -96; c = -27;
Δ = b2-4ac
Δ = -962-4·16·(-27)
Δ = 10944
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:

$\sqrt{\Delta}=\sqrt{10944}=\sqrt{576*19}=\sqrt{576}*\sqrt{19}=24\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-24\sqrt{19}}{2*16}=\frac{96-24\sqrt{19}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+24\sqrt{19}}{2*16}=\frac{96+24\sqrt{19}}{32}
$