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3b^2-17b+10=0
a = 3; b = -17; c = +10;
Δ = b2-4ac
Δ = -172-4·3·10
Δ = 169
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{169}=13$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-13}{2*3}=\frac{4}{6} =2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+13}{2*3}=\frac{30}{6} =5 $
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