0=-16t^2+160t+20

Solution for 0=-16t^2+160t+20 equation:

0=-16t^2+160t+20

We move all terms to the left:

0-(-16t^2+160t+20)=0

We add all the numbers together, and all the variables

-(-16t^2+160t+20)=0

We get rid of parentheses

16t^2-160t-20=0

a = 16; b = -160; c = -20;
Δ = b2-4ac
Δ = -1602-4·16·(-20)
Δ = 26880
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:

$\sqrt{\Delta}=\sqrt{26880}=\sqrt{256*105}=\sqrt{256}*\sqrt{105}=16\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-16\sqrt{105}}{2*16}=\frac{160-16\sqrt{105}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+16\sqrt{105}}{2*16}=\frac{160+16\sqrt{105}}{32}
$