20d^2+41d+2=0

Solution for 20d^2+41d+2=0 equation:

20d^2+41d+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:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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