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7k^2-19k-9=0
a = 7; b = -19; c = -9;
Δ = b2-4ac
Δ = -192-4·7·(-9)
Δ = 613
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{-(-19)-\sqrt{613}}{2*7}=\frac{19-\sqrt{613}}{14} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{613}}{2*7}=\frac{19+\sqrt{613}}{14} $
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