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6k^2-36k=0
a = 6; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·6·0
Δ = 1296
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{1296}=36$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*6}=\frac{0}{12} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*6}=\frac{72}{12} =6 $
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