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3k^2-18k+15=0
a = 3; b = -18; c = +15;
Δ = b2-4ac
Δ = -182-4·3·15
Δ = 144
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{144}=12$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12}{2*3}=\frac{6}{6} =1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12}{2*3}=\frac{30}{6} =5 $
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