0=-16t^2+33t+9

Solution for 0=-16t^2+33t+9 equation:

0=-16t^2+33t+9

We move all terms to the left:

0-(-16t^2+33t+9)=0

We add all the numbers together, and all the variables

-(-16t^2+33t+9)=0

We get rid of parentheses

16t^2-33t-9=0

a = 16; b = -33; c = -9;
Δ = b2-4ac
Δ = -332-4·16·(-9)
Δ = 1665
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:

$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3\sqrt{185}}{2*16}=\frac{33-3\sqrt{185}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3\sqrt{185}}{2*16}=\frac{33+3\sqrt{185}}{32}
$