3j2+9j+4=0

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Solution for 3j2+9j+4=0 equation: 



3j^2+9j+4=0

a = 3; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·3·4
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{33}}{2*3}=\frac{-9-\sqrt{33}}{6}
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{33}}{2*3}=\frac{-9+\sqrt{33}}{6}
$

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