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

Solution for -5-10t+20t^2=0 equation:

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

a = 20; b = -10; c = -5;
Δ = b2-4ac
Δ = -102-4·20·(-5)
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{-(-10)-10\sqrt{5}}{2*20}=\frac{10-10\sqrt{5}}{40}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{5}}{2*20}=\frac{10+10\sqrt{5}}{40}
$