-5t2+20t+10=0

Solution for -5t2+20t+10=0 equation:

-5t^2+20t+10=0

a = -5; b = 20; c = +10;
Δ = b2-4ac
Δ = 202-4·(-5)·10
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{6}}{2*-5}=\frac{-20-10\sqrt{6}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{6}}{2*-5}=\frac{-20+10\sqrt{6}}{-10}
$