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7b^2+19b-6=0
a = 7; b = 19; c = -6;
Δ = b2-4ac
Δ = 192-4·7·(-6)
Δ = 529
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{529}=23$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-23}{2*7}=\frac{-42}{14} =-3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*7}=\frac{4}{14} =2/7 $
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