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6j^2-9j-8=0
a = 6; b = -9; c = -8;
Δ = b2-4ac
Δ = -92-4·6·(-8)
Δ = 273
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{273}}{2*6}=\frac{9-\sqrt{273}}{12} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{273}}{2*6}=\frac{9+\sqrt{273}}{12} $
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