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r^2-41r+400=0
a = 1; b = -41; c = +400;
Δ = b2-4ac
Δ = -412-4·1·400
Δ = 81
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{81}=9$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-9}{2*1}=\frac{32}{2} =16 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+9}{2*1}=\frac{50}{2} =25 $
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