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3k-18k^2=0
a = -18; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-18)·0
Δ = 9
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{9}=3$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-18}=\frac{-6}{-36} =1/6 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-18}=\frac{0}{-36} =0 $
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