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9k^2-24=3k
We move all terms to the left:
9k^2-24-(3k)=0
a = 9; b = -3; c = -24;
Δ = b2-4ac
Δ = -32-4·9·(-24)
Δ = 873
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{97}}{2*9}=\frac{3-3\sqrt{97}}{18} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{97}}{2*9}=\frac{3+3\sqrt{97}}{18} $
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