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6b^2+13b-20=0
a = 6; b = 13; c = -20;
Δ = b2-4ac
Δ = 132-4·6·(-20)
Δ = 649
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{-(13)-\sqrt{649}}{2*6}=\frac{-13-\sqrt{649}}{12} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{649}}{2*6}=\frac{-13+\sqrt{649}}{12} $
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