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3k^2+k=6
We move all terms to the left:
3k^2+k-(6)=0
a = 3; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·3·(-6)
Δ = 73
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{73}}{2*3}=\frac{-1-\sqrt{73}}{6} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{73}}{2*3}=\frac{-1+\sqrt{73}}{6} $
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