20=-16t^2+50t

Solution for 20=-16t^2+50t equation:

20=-16t^2+50t

We move all terms to the left:

20-(-16t^2+50t)=0

We get rid of parentheses

16t^2-50t+20=0

a = 16; b = -50; c = +20;
Δ = b2-4ac
Δ = -502-4·16·20
Δ = 1220
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:

$\sqrt{\Delta}=\sqrt{1220}=\sqrt{4*305}=\sqrt{4}*\sqrt{305}=2\sqrt{305}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{305}}{2*16}=\frac{50-2\sqrt{305}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{305}}{2*16}=\frac{50+2\sqrt{305}}{32}
$