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9k^2+k=0
a = 9; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·9·0
Δ = 1
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{1}=1$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*9}=\frac{-2}{18} =-1/9 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*9}=\frac{0}{18} =0 $
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