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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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-12}=\frac{-8}{-24} =1/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-12}=\frac{0}{-24} =0 $
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