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6b^2+11b-7=0
a = 6; b = 11; c = -7;
Δ = b2-4ac
Δ = 112-4·6·(-7)
Δ = 289
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{289}=17$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-17}{2*6}=\frac{-28}{12} =-2+1/3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+17}{2*6}=\frac{6}{12} =1/2 $
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