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2k^2+19k-30=0
a = 2; b = 19; c = -30;
Δ = b2-4ac
Δ = 192-4·2·(-30)
Δ = 601
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{601}}{2*2}=\frac{-19-\sqrt{601}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{601}}{2*2}=\frac{-19+\sqrt{601}}{4} $
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