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0=10r-3r^2
We move all terms to the left:
0-(10r-3r^2)=0
We add all the numbers together, and all the variables
-(10r-3r^2)=0
We get rid of parentheses
3r^2-10r=0
a = 3; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·3·0
Δ = 100
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{100}=10$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*3}=\frac{0}{6} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*3}=\frac{20}{6} =3+1/3 $
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