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7k^2+k-13=0
a = 7; b = 1; c = -13;
Δ = b2-4ac
Δ = 12-4·7·(-13)
Δ = 365
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{365}}{2*7}=\frac{-1-\sqrt{365}}{14} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{365}}{2*7}=\frac{-1+\sqrt{365}}{14} $
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