-5t^2+100t-80=0

Solution for -5t^2+100t-80=0 equation:

-5t^2+100t-80=0

a = -5; b = 100; c = -80;
Δ = b2-4ac
Δ = 1002-4·(-5)·(-80)
Δ = 8400
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:

$\sqrt{\Delta}=\sqrt{8400}=\sqrt{400*21}=\sqrt{400}*\sqrt{21}=20\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20\sqrt{21}}{2*-5}=\frac{-100-20\sqrt{21}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20\sqrt{21}}{2*-5}=\frac{-100+20\sqrt{21}}{-10}
$