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=-2R^2+130R
We move all terms to the left:
-(-2R^2+130R)=0
We get rid of parentheses
2R^2-130R=0
a = 2; b = -130; c = 0;
Δ = b2-4ac
Δ = -1302-4·2·0
Δ = 16900
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{16900}=130$$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-130)-130}{2*2}=\frac{0}{4} =0 $$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-130)+130}{2*2}=\frac{260}{4} =65 $
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