3j^2+8j-3=0

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Solution for 3j^2+8j-3=0 equation: 



3j^2+8j-3=0

a = 3; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·3·(-3)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-10}{2*3}=\frac{-18}{6}
=-3
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+10}{2*3}=\frac{2}{6}
=1/3
$

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