-16t^2+33t=0

Solution for -16t^2+33t=0 equation:

-16t^2+33t=0

a = -16; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·(-16)·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*-16}=\frac{-66}{-32}
=2+1/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*-16}=\frac{0}{-32}
=0
$