0=-16t^2+80t-64

Solution for 0=-16t^2+80t-64 equation:

0=-16t^2+80t-64

We move all terms to the left:

0-(-16t^2+80t-64)=0

We add all the numbers together, and all the variables

-(-16t^2+80t-64)=0

We get rid of parentheses

16t^2-80t+64=0

a = 16; b = -80; c = +64;
Δ = b2-4ac
Δ = -802-4·16·64
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-48}{2*16}=\frac{32}{32}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+48}{2*16}=\frac{128}{32}
=4
$