0=16t^2+40t+20

Solution for 0=16t^2+40t+20 equation:

0=16t^2+40t+20

We move all terms to the left:

0-(16t^2+40t+20)=0

We add all the numbers together, and all the variables

-(16t^2+40t+20)=0

We get rid of parentheses

-16t^2-40t-20=0

a = -16; b = -40; c = -20;
Δ = b2-4ac
Δ = -402-4·(-16)·(-20)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{5}}{2*-16}=\frac{40-8\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{5}}{2*-16}=\frac{40+8\sqrt{5}}{-32}
$