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6j^2-96=0
a = 6; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·6·(-96)
Δ = 2304
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{2304}=48$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*6}=\frac{-48}{12} =-4 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*6}=\frac{48}{12} =4 $
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