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r2+12r=0
We add all the numbers together, and all the variables
r^2+12r=0
a = 1; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·1·0
Δ = 144
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{144}=12$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*1}=\frac{-24}{2} =-12 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*1}=\frac{0}{2} =0 $
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