6j^2+j-8=0

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Solution for 6j^2+j-8=0 equation: 



6j^2+j-8=0

a = 6; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·6·(-8)
Δ = 193
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{-(1)-\sqrt{193}}{2*6}=\frac{-1-\sqrt{193}}{12}
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{193}}{2*6}=\frac{-1+\sqrt{193}}{12}
$

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