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9k^2+21k-18=0
a = 9; b = 21; c = -18;
Δ = b2-4ac
Δ = 212-4·9·(-18)
Δ = 1089
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{1089}=33$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-33}{2*9}=\frac{-54}{18} =-3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+33}{2*9}=\frac{12}{18} =2/3 $
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