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6j^2+29j-5=0
a = 6; b = 29; c = -5;
Δ = b2-4ac
Δ = 292-4·6·(-5)
Δ = 961
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{961}=31$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-31}{2*6}=\frac{-60}{12} =-5 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+31}{2*6}=\frac{2}{12} =1/6 $
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