-5t^2-20t+80=0

Solution for -5t^2-20t+80=0 equation:

-5t^2-20t+80=0

a = -5; b = -20; c = +80;
Δ = b2-4ac
Δ = -202-4·(-5)·80
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{5}}{2*-5}=\frac{20-20\sqrt{5}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{5}}{2*-5}=\frac{20+20\sqrt{5}}{-10}
$