-5t^2+20t+1,6=0

Solution for -5t^2+20t+1,6=0 equation:

-5t^2+20t+1.6=0

a = -5; b = 20; c = +1.6;
Δ = b2-4ac
Δ = 202-4·(-5)·1.6
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{3}}{2*-5}=\frac{-20-12\sqrt{3}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{3}}{2*-5}=\frac{-20+12\sqrt{3}}{-10}
$