0=16t^2+80t+20

Solution for 0=16t^2+80t+20 equation:

0=16t^2+80t+20

We move all terms to the left:

0-(16t^2+80t+20)=0

We add all the numbers together, and all the variables

-(16t^2+80t+20)=0

We get rid of parentheses

-16t^2-80t-20=0

a = -16; b = -80; c = -20;
Δ = b2-4ac
Δ = -802-4·(-16)·(-20)
Δ = 5120
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:

$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-32\sqrt{5}}{2*-16}=\frac{80-32\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+32\sqrt{5}}{2*-16}=\frac{80+32\sqrt{5}}{-32}
$