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2j^2-19j+9=0
a = 2; b = -19; c = +9;
Δ = b2-4ac
Δ = -192-4·2·9
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$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-17}{2*2}=\frac{2}{4} =1/2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+17}{2*2}=\frac{36}{4} =9 $
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