1.6t^2-20t=0

Solution for 1.6t^2-20t=0 equation:

1.6t^2-20t=0

a = 1.6; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·1.6·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*1.6}=\frac{0}{3.2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*1.6}=\frac{40}{3.2}
=12+1/2
$