-5t^2+40t-20=0

Solution for -5t^2+40t-20=0 equation:

-5t^2+40t-20=0

a = -5; b = 40; c = -20;
Δ = b2-4ac
Δ = 402-4·(-5)·(-20)
Δ = 1200
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:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{3}}{2*-5}=\frac{-40-20\sqrt{3}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{3}}{2*-5}=\frac{-40+20\sqrt{3}}{-10}
$