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10r^2=25r
We move all terms to the left:
10r^2-(25r)=0
a = 10; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·10·0
Δ = 625
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{625}=25$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*10}=\frac{0}{20} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*10}=\frac{50}{20} =2+1/2 $
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