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3b^2+6-19b=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:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-17}{2*3}=\frac{2}{6} =1/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+17}{2*3}=\frac{36}{6} =6 $
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