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

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

1.6t^2-20t+1=0

a = 1.6; b = -20; c = +1;
Δ = b2-4ac
Δ = -202-4·1.6·1
Δ = 393.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-\sqrt{393.6}}{2*1.6}=\frac{20-\sqrt{393.6}}{3.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{393.6}}{2*1.6}=\frac{20+\sqrt{393.6}}{3.2}
$