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3b^2-17b+15=0
a = 3; b = -17; c = +15;
Δ = b2-4ac
Δ = -172-4·3·15
Δ = 109
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{-(-17)-\sqrt{109}}{2*3}=\frac{17-\sqrt{109}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{109}}{2*3}=\frac{17+\sqrt{109}}{6} $
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