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3r^2=4r-1
We move all terms to the left:
3r^2-(4r-1)=0
We get rid of parentheses
3r^2-4r+1=0
a = 3; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·3·1
Δ = 4
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{4}=2$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*3}=\frac{2}{6} =1/3 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*3}=\frac{6}{6} =1 $
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