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14x^2+x-13=0
a = 14; b = 1; c = -13;
Δ = b2-4ac
Δ = 12-4·14·(-13)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-27}{2*14}=\frac{-28}{28} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+27}{2*14}=\frac{26}{28} =13/14 $
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