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9j^2+39j+12=0
a = 9; b = 39; c = +12;
Δ = b2-4ac
Δ = 392-4·9·12
Δ = 1089
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{1089}=33$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-33}{2*9}=\frac{-72}{18} =-4 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+33}{2*9}=\frac{-6}{18} =-1/3 $
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