20=-16t^2+56t

Solution for 20=-16t^2+56t equation:

20=-16t^2+56t

We move all terms to the left:

20-(-16t^2+56t)=0

We get rid of parentheses

16t^2-56t+20=0

a = 16; b = -56; c = +20;
Δ = b2-4ac
Δ = -562-4·16·20
Δ = 1856
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:

$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-8\sqrt{29}}{2*16}=\frac{56-8\sqrt{29}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+8\sqrt{29}}{2*16}=\frac{56+8\sqrt{29}}{32}
$