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=289R-R^2
We move all terms to the left:
-(289R-R^2)=0
We get rid of parentheses
R^2-289R=0
a = 1; b = -289; c = 0;
Δ = b2-4ac
Δ = -2892-4·1·0
Δ = 83521
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{83521}=289$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-289)-289}{2*1}=\frac{0}{2} =0 $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-289)+289}{2*1}=\frac{578}{2} =289 $
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