20x2+9x+1=0

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Solution for 20x2+9x+1=0 equation: 



20x^2+9x+1=0

a = 20; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·20·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*20}=\frac{-10}{40}
=-1/4
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*20}=\frac{-8}{40}
=-1/5
$

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