0=-16t^2+24t+432

Solution for 0=-16t^2+24t+432 equation:

0=-16t^2+24t+432

We move all terms to the left:

0-(-16t^2+24t+432)=0

We add all the numbers together, and all the variables

-(-16t^2+24t+432)=0

We get rid of parentheses

16t^2-24t-432=0

a = 16; b = -24; c = -432;
Δ = b2-4ac
Δ = -242-4·16·(-432)
Δ = 28224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28224}=168$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-168}{2*16}=\frac{-144}{32}
=-4+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+168}{2*16}=\frac{192}{32}
=6
$