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6k^2-12k-18=0
a = 6; b = -12; c = -18;
Δ = b2-4ac
Δ = -122-4·6·(-18)
Δ = 576
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{576}=24$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-24}{2*6}=\frac{-12}{12} =-1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+24}{2*6}=\frac{36}{12} =3 $
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