32+32t-16t^2=0

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

32+32t-16t^2=0

a = -16; b = 32; c = +32;
Δ = b2-4ac
Δ = 322-4·(-16)·32
Δ = 3072
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:

$\sqrt{\Delta}=\sqrt{3072}=\sqrt{1024*3}=\sqrt{1024}*\sqrt{3}=32\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{3}}{2*-16}=\frac{-32-32\sqrt{3}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{3}}{2*-16}=\frac{-32+32\sqrt{3}}{-32}
$