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6j^2-57j-30=0
a = 6; b = -57; c = -30;
Δ = b2-4ac
Δ = -572-4·6·(-30)
Δ = 3969
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{3969}=63$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-57)-63}{2*6}=\frac{-6}{12} =-1/2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-57)+63}{2*6}=\frac{120}{12} =10 $
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