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13r=r^2
We move all terms to the left:
13r-(r^2)=0
determiningTheFunctionDomain -r^2+13r=0
We add all the numbers together, and all the variables
-1r^2+13r=0
a = -1; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-1)·0
Δ = 169
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{169}=13$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-1}=\frac{-26}{-2} =+13 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-1}=\frac{0}{-2} =0 $
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