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8/3r-9/4r=-52/12
We move all terms to the left:
8/3r-9/4r-(-52/12)=0
Domain of the equation: 3r!=0
r!=0/3
r!=0
r∈R
Domain of the equation: 4r!=0We get rid of parentheses
r!=0/4
r!=0
r∈R
8/3r-9/4r+52/12=0
We calculate fractions
2496r^2/144r^2+384r/144r^2+(-324r)/144r^2=0
We multiply all the terms by the denominator
2496r^2+384r+(-324r)=0
We get rid of parentheses
2496r^2+384r-324r=0
We add all the numbers together, and all the variables
2496r^2+60r=0
a = 2496; b = 60; c = 0;
Δ = b2-4ac
Δ = 602-4·2496·0
Δ = 3600
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{3600}=60$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60}{2*2496}=\frac{-120}{4992} =-5/208 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60}{2*2496}=\frac{0}{4992} =0 $
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