-16t^2+96+432=0

Solution for -16t^2+96+432=0 equation:

-16t^2+96+432=0

We add all the numbers together, and all the variables

-16t^2+528=0

a = -16; b = 0; c = +528;
Δ = b2-4ac
Δ = 02-4·(-16)·528
Δ = 33792
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:

$\sqrt{\Delta}=\sqrt{33792}=\sqrt{1024*33}=\sqrt{1024}*\sqrt{33}=32\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{33}}{2*-16}=\frac{0-32\sqrt{33}}{-32}
=-\frac{32\sqrt{33}}{-32}
=-\frac{\sqrt{33}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{33}}{2*-16}=\frac{0+32\sqrt{33}}{-32}
=\frac{32\sqrt{33}}{-32}
=\frac{\sqrt{33}}{-1}
$