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2r^2-18r=0
a = 2; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·2·0
Δ = 324
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{324}=18$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*2}=\frac{0}{4} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*2}=\frac{36}{4} =9 $
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