-16t^2+288t+32=0

Solution for -16t^2+288t+32=0 equation:

-16t^2+288t+32=0

a = -16; b = 288; c = +32;
Δ = b2-4ac
Δ = 2882-4·(-16)·32
Δ = 84992
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:

$\sqrt{\Delta}=\sqrt{84992}=\sqrt{1024*83}=\sqrt{1024}*\sqrt{83}=32\sqrt{83}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(288)-32\sqrt{83}}{2*-16}=\frac{-288-32\sqrt{83}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(288)+32\sqrt{83}}{2*-16}=\frac{-288+32\sqrt{83}}{-32}
$