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-3j^2-9j-5=0
a = -3; b = -9; c = -5;
Δ = b2-4ac
Δ = -92-4·(-3)·(-5)
Δ = 21
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}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{21}}{2*-3}=\frac{9-\sqrt{21}}{-6} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{21}}{2*-3}=\frac{9+\sqrt{21}}{-6} $
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