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20r^2=13r+15
We move all terms to the left:
20r^2-(13r+15)=0
We get rid of parentheses
20r^2-13r-15=0
a = 20; b = -13; c = -15;
Δ = b2-4ac
Δ = -132-4·20·(-15)
Δ = 1369
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{1369}=37$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-37}{2*20}=\frac{-24}{40} =-3/5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+37}{2*20}=\frac{50}{40} =1+1/4 $
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