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1r-3=3(4-3/2r)
We move all terms to the left:
1r-3-(3(4-3/2r))=0
Domain of the equation: 2r))!=0We add all the numbers together, and all the variables
r!=0/1
r!=0
r∈R
1r-(3(-3/2r+4))-3=0
We add all the numbers together, and all the variables
r-(3(-3/2r+4))-3=0
We multiply all the terms by the denominator
r*2r-3*2r-3+4))-(3(+4))=0
We add all the numbers together, and all the variables
r*2r-3*2r-3+4))-(34)=0
We add all the numbers together, and all the variables
r*2r-3*2r=0
Wy multiply elements
2r^2-6r=0
a = 2; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·2·0
Δ = 36
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{36}=6$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*2}=\frac{0}{4} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*2}=\frac{12}{4} =3 $
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