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9b^2+18b+5=0
a = 9; b = 18; c = +5;
Δ = b2-4ac
Δ = 182-4·9·5
Δ = 144
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{144}=12$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12}{2*9}=\frac{-30}{18} =-1+2/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12}{2*9}=\frac{-6}{18} =-1/3 $
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