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(R)=500R-0.4R^2
We move all terms to the left:
(R)-(500R-0.4R^2)=0
We get rid of parentheses
0.4R^2-500R+R=0
We add all the numbers together, and all the variables
0.4R^2-499R=0
a = 0.4; b = -499; c = 0;
Δ = b2-4ac
Δ = -4992-4·0.4·0
Δ = 249001
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{249001}=499$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-499)-499}{2*0.4}=\frac{0}{0.8} =0 $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-499)+499}{2*0.4}=\frac{998}{0.8} =1247+0.4/0.8 $
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