0=-500t^2+80t+10

Solution for 0=-500t^2+80t+10 equation:

0=-500t^2+80t+10

We move all terms to the left:

0-(-500t^2+80t+10)=0

We add all the numbers together, and all the variables

-(-500t^2+80t+10)=0

We get rid of parentheses

500t^2-80t-10=0

a = 500; b = -80; c = -10;
Δ = b2-4ac
Δ = -802-4·500·(-10)
Δ = 26400
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:

$\sqrt{\Delta}=\sqrt{26400}=\sqrt{400*66}=\sqrt{400}*\sqrt{66}=20\sqrt{66}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-20\sqrt{66}}{2*500}=\frac{80-20\sqrt{66}}{1000}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+20\sqrt{66}}{2*500}=\frac{80+20\sqrt{66}}{1000}
$