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6b^2+13b+5=0
a = 6; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·6·5
Δ = 49
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{49}=7$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*6}=\frac{-20}{12} =-1+2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*6}=\frac{-6}{12} =-1/2 $
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