0=20+32t+-16t2

Solution for 0=20+32t+-16t2 equation:

0=20+32t+-16t^2

We move all terms to the left:

0-(20+32t+-16t^2)=0

We add all the numbers together, and all the variables

-(20+32t+-16t^2)=0

We use the square of the difference formula

-(20+32t-16t^2)=0

We get rid of parentheses

16t^2-32t-20=0

a = 16; b = -32; c = -20;
Δ = b2-4ac
Δ = -322-4·16·(-20)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-48}{2*16}=\frac{-16}{32}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+48}{2*16}=\frac{80}{32}
=2+1/2
$