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3k^2-18k-21=0
a = 3; b = -18; c = -21;
Δ = b2-4ac
Δ = -182-4·3·(-21)
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-24}{2*3}=\frac{-6}{6} =-1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+24}{2*3}=\frac{42}{6} =7 $
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