16t^2-40t+20=0

Solution for 16t^2-40t+20=0 equation:

16t^2-40t+20=0

a = 16; b = -40; c = +20;
Δ = b2-4ac
Δ = -402-4·16·20
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{5}}{2*16}=\frac{40-8\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{5}}{2*16}=\frac{40+8\sqrt{5}}{32}
$