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20r^2-12r=0
a = 20; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·20·0
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$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*20}=\frac{0}{40} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*20}=\frac{24}{40} =3/5 $
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