20x2+39x+19=0

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Solution for 20x2+39x+19=0 equation:



20x^2+39x+19=0
a = 20; b = 39; c = +19;
Δ = b2-4ac
Δ = 392-4·20·19
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{-(39)-1}{2*20}=\frac{-40}{40} =-1 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+1}{2*20}=\frac{-38}{40} =-19/20 $

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