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5b^2+b2+9b=2
We move all terms to the left:
5b^2+b2+9b-(2)=0
We add all the numbers together, and all the variables
6b^2+9b-2=0
a = 6; b = 9; c = -2;
Δ = b2-4ac
Δ = 92-4·6·(-2)
Δ = 129
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{129}}{2*6}=\frac{-9-\sqrt{129}}{12} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{129}}{2*6}=\frac{-9+\sqrt{129}}{12} $
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