20x^2+-41x+-9=0

Solution for 20x^2+-41x+-9=0 equation:

20x^2+-41x+-9=0

We add all the numbers together, and all the variables

20x^2-41x=0

a = 20; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·20·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*20}=\frac{0}{40}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*20}=\frac{82}{40}
=2+1/20
$