0=-16t^2+107t+63

Solution for 0=-16t^2+107t+63 equation:

0=-16t^2+107t+63

We move all terms to the left:

0-(-16t^2+107t+63)=0

We add all the numbers together, and all the variables

-(-16t^2+107t+63)=0

We get rid of parentheses

16t^2-107t-63=0

a = 16; b = -107; c = -63;
Δ = b2-4ac
Δ = -1072-4·16·(-63)
Δ = 15481
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{-(-107)-\sqrt{15481}}{2*16}=\frac{107-\sqrt{15481}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-107)+\sqrt{15481}}{2*16}=\frac{107+\sqrt{15481}}{32}
$