20=48t-16t^2

Solution for 20=48t-16t^2 equation:

20=48t-16t^2

We move all terms to the left:

20-(48t-16t^2)=0

We get rid of parentheses

16t^2-48t+20=0

a = 16; b = -48; c = +20;
Δ = b2-4ac
Δ = -482-4·16·20
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32}{2*16}=\frac{16}{32}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32}{2*16}=\frac{80}{32}
=2+1/2
$