20=-16t^2+63t+4

Solution for 20=-16t^2+63t+4 equation:

20=-16t^2+63t+4

We move all terms to the left:

20-(-16t^2+63t+4)=0

We get rid of parentheses

16t^2-63t-4+20=0

We add all the numbers together, and all the variables

16t^2-63t+16=0

a = 16; b = -63; c = +16;
Δ = b2-4ac
Δ = -632-4·16·16
Δ = 2945
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{2945}}{2*16}=\frac{63-\sqrt{2945}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-63)+\sqrt{2945}}{2*16}=\frac{63+\sqrt{2945}}{32}
$