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(R)=123.5R-R^2
We move all terms to the left:
(R)-(123.5R-R^2)=0
We get rid of parentheses
R^2-123.5R+R=0
We add all the numbers together, and all the variables
R^2-122.5R=0
a = 1; b = -122.5; c = 0;
Δ = b2-4ac
Δ = -122.52-4·1·0
Δ = 15006.25
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{-(-122.5)-\sqrt{15006.25}}{2*1}=\frac{122.5-\sqrt{15006.25}}{2} $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-122.5)+\sqrt{15006.25}}{2*1}=\frac{122.5+\sqrt{15006.25}}{2} $
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