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(R)=-R^2+12R
We move all terms to the left:
(R)-(-R^2+12R)=0
We get rid of parentheses
R^2-12R+R=0
We add all the numbers together, and all the variables
R^2-11R=0
a = 1; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·1·0
Δ = 121
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{121}=11$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*1}=\frac{0}{2} =0 $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*1}=\frac{22}{2} =11 $
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