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12k^2-4k=0
a = 12; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·12·0
Δ = 16
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{16}=4$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*12}=\frac{0}{24} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*12}=\frac{8}{24} =1/3 $
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