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

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

-32t-16t^2+32=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}
$