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3k^2-24k=0
a = 3; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·3·0
Δ = 576
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{576}=24$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*3}=\frac{0}{6} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*3}=\frac{48}{6} =8 $
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