-10t^2+50t+20=0

Solution for -10t^2+50t+20=0 equation:

-10t^2+50t+20=0

a = -10; b = 50; c = +20;
Δ = b2-4ac
Δ = 502-4·(-10)·20
Δ = 3300
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:

$\sqrt{\Delta}=\sqrt{3300}=\sqrt{100*33}=\sqrt{100}*\sqrt{33}=10\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{33}}{2*-10}=\frac{-50-10\sqrt{33}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{33}}{2*-10}=\frac{-50+10\sqrt{33}}{-20}
$