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r^2-2r+0.36=0
a = 1; b = -2; c = +0.36;
Δ = b2-4ac
Δ = -22-4·1·0.36
Δ = 2.56
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-\sqrt{2.56}}{2*1}=\frac{2-\sqrt{2.56}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{2.56}}{2*1}=\frac{2+\sqrt{2.56}}{2} $
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