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3j^2+19j+6=0
a = 3; b = 19; c = +6;
Δ = b2-4ac
Δ = 192-4·3·6
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$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*3}=\frac{-36}{6} =-6 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*3}=\frac{-2}{6} =-1/3 $
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