20t^2-41t+20=0

Solution for 20t^2-41t+20=0 equation:

20t^2-41t+20=0

a = 20; b = -41; c = +20;
Δ = b2-4ac
Δ = -412-4·20·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-9}{2*20}=\frac{32}{40}
=4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+9}{2*20}=\frac{50}{40}
=1+1/4
$