0=-16t^2+98t+33

Solution for 0=-16t^2+98t+33 equation:

0=-16t^2+98t+33

We move all terms to the left:

0-(-16t^2+98t+33)=0

We add all the numbers together, and all the variables

-(-16t^2+98t+33)=0

We get rid of parentheses

16t^2-98t-33=0

a = 16; b = -98; c = -33;
Δ = b2-4ac
Δ = -982-4·16·(-33)
Δ = 11716
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:

$\sqrt{\Delta}=\sqrt{11716}=\sqrt{4*2929}=\sqrt{4}*\sqrt{2929}=2\sqrt{2929}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-2\sqrt{2929}}{2*16}=\frac{98-2\sqrt{2929}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+2\sqrt{2929}}{2*16}=\frac{98+2\sqrt{2929}}{32}
$