0=-16t^2+63t+17

Solution for 0=-16t^2+63t+17 equation:

0=-16t^2+63t+17

We move all terms to the left:

0-(-16t^2+63t+17)=0

We add all the numbers together, and all the variables

-(-16t^2+63t+17)=0

We get rid of parentheses

16t^2-63t-17=0

a = 16; b = -63; c = -17;
Δ = b2-4ac
Δ = -632-4·16·(-17)
Δ = 5057
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-63)-\sqrt{5057}}{2*16}=\frac{63-\sqrt{5057}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+\sqrt{5057}}{2*16}=\frac{63+\sqrt{5057}}{32}
$