19j2+9j=0

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Solution for 19j2+9j=0 equation: 



19j^2+9j=0

a = 19; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·19·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*19}=\frac{-18}{38}
=-9/19
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*19}=\frac{0}{38}
=0
$

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