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9r^2+24r=0
a = 9; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·9·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*9}=\frac{-48}{18} =-2+2/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*9}=\frac{0}{18} =0 $
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