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14b^2-3b-2=0
a = 14; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·14·(-2)
Δ = 121
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{121}=11$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*14}=\frac{-8}{28} =-2/7 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*14}=\frac{14}{28} =1/2 $
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