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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16t^2-159t-20=0

a = 16; b = -159; c = -20;
Δ = b2-4ac
Δ = -1592-4·16·(-20)
Δ = 26561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-159)-\sqrt{26561}}{2*16}=\frac{159-\sqrt{26561}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-159)+\sqrt{26561}}{2*16}=\frac{159+\sqrt{26561}}{32}
$