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8j^2+15j-2=0
a = 8; b = 15; c = -2;
Δ = b2-4ac
Δ = 152-4·8·(-2)
Δ = 289
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{289}=17$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-17}{2*8}=\frac{-32}{16} =-2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+17}{2*8}=\frac{2}{16} =1/8 $
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