-5t^2+20t-8=0

Solution for -5t^2+20t-8=0 equation:

-5t^2+20t-8=0

a = -5; b = 20; c = -8;
Δ = b2-4ac
Δ = 202-4·(-5)·(-8)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*-5}=\frac{-20-4\sqrt{15}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*-5}=\frac{-20+4\sqrt{15}}{-10}
$