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6j^2-35j-6=0
a = 6; b = -35; c = -6;
Δ = b2-4ac
Δ = -352-4·6·(-6)
Δ = 1369
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{1369}=37$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-37}{2*6}=\frac{-2}{12} =-1/6 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+37}{2*6}=\frac{72}{12} =6 $
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