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6k^2-23k-4=0
a = 6; b = -23; c = -4;
Δ = b2-4ac
Δ = -232-4·6·(-4)
Δ = 625
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-25}{2*6}=\frac{-2}{12} =-1/6 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+25}{2*6}=\frac{48}{12} =4 $
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