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r=4/3r-5
We move all terms to the left:
r-(4/3r-5)=0
Domain of the equation: 3r-5)!=0We get rid of parentheses
r∈R
r-4/3r+5=0
We multiply all the terms by the denominator
r*3r+5*3r-4=0
Wy multiply elements
3r^2+15r-4=0
a = 3; b = 15; c = -4;
Δ = b2-4ac
Δ = 152-4·3·(-4)
Δ = 273
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{273}}{2*3}=\frac{-15-\sqrt{273}}{6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{273}}{2*3}=\frac{-15+\sqrt{273}}{6} $
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