20v^2-41v+2=0

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

20v^2-41v+2=0

a = 20; b = -41; c = +2;
Δ = b2-4ac
Δ = -412-4·20·2
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-39}{2*20}=\frac{2}{40}
=1/20
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+39}{2*20}=\frac{80}{40}
=2
$