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3j^2-49j+16=0
a = 3; b = -49; c = +16;
Δ = b2-4ac
Δ = -492-4·3·16
Δ = 2209
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{2209}=47$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-47}{2*3}=\frac{2}{6} =1/3 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+47}{2*3}=\frac{96}{6} =16 $
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