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3b^2-9b-17=0
a = 3; b = -9; c = -17;
Δ = b2-4ac
Δ = -92-4·3·(-17)
Δ = 285
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{285}}{2*3}=\frac{9-\sqrt{285}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{285}}{2*3}=\frac{9+\sqrt{285}}{6} $
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