0=-16t^2+56t-33

Solution for 0=-16t^2+56t-33 equation:

0=-16t^2+56t-33

We move all terms to the left:

0-(-16t^2+56t-33)=0

We add all the numbers together, and all the variables

-(-16t^2+56t-33)=0

We get rid of parentheses

16t^2-56t+33=0

a = 16; b = -56; c = +33;
Δ = b2-4ac
Δ = -562-4·16·33
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-32}{2*16}=\frac{24}{32}
=3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+32}{2*16}=\frac{88}{32}
=2+3/4
$