0=-10t^2+80

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

0=-10t^2+80

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

10t^2-80=0

a = 10; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·10·(-80)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*10}=\frac{0-40\sqrt{2}}{20}
=-\frac{40\sqrt{2}}{20}
=-2\sqrt{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*10}=\frac{0+40\sqrt{2}}{20}
=\frac{40\sqrt{2}}{20}
=2\sqrt{2}
$