0=-16t^2+63

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

0=-16t^2+63

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-63=0

a = 16; b = 0; c = -63;
Δ = b2-4ac
Δ = 02-4·16·(-63)
Δ = 4032
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:

$\sqrt{\Delta}=\sqrt{4032}=\sqrt{576*7}=\sqrt{576}*\sqrt{7}=24\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{7}}{2*16}=\frac{0-24\sqrt{7}}{32}
=-\frac{24\sqrt{7}}{32}
=-\frac{3\sqrt{7}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{7}}{2*16}=\frac{0+24\sqrt{7}}{32}
=\frac{24\sqrt{7}}{32}
=\frac{3\sqrt{7}}{4}
$