0=16t^2+32t=20

Solution for 0=16t^2+32t=20 equation:

0=16t^2+32t=20

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-16t^2-32t=0

a = -16; b = -32; c = 0;
Δ = b2-4ac
Δ = -322-4·(-16)·0
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{-(-32)-32}{2*-16}=\frac{0}{-32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32}{2*-16}=\frac{64}{-32}
=-2
$