-5t^2+30t-20=0

Solution for -5t^2+30t-20=0 equation:

-5t^2+30t-20=0

a = -5; b = 30; c = -20;
Δ = b2-4ac
Δ = 302-4·(-5)·(-20)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-10\sqrt{5}}{2*-5}=\frac{-30-10\sqrt{5}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+10\sqrt{5}}{2*-5}=\frac{-30+10\sqrt{5}}{-10}
$