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r=2/3r
We move all terms to the left:
r-(2/3r)=0
Domain of the equation: 3r)!=0We add all the numbers together, and all the variables
r!=0/1
r!=0
r∈R
r-(+2/3r)=0
We get rid of parentheses
r-2/3r=0
We multiply all the terms by the denominator
r*3r-2=0
Wy multiply elements
3r^2-2=0
a = 3; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·3·(-2)
Δ = 24
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*3}=\frac{0-2\sqrt{6}}{6} =-\frac{2\sqrt{6}}{6} =-\frac{\sqrt{6}}{3} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*3}=\frac{0+2\sqrt{6}}{6} =\frac{2\sqrt{6}}{6} =\frac{\sqrt{6}}{3} $
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