(r^2+5r-6)/4r+(3/2r)=r-4

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Solution for (r^2+5r-6)/4r+(3/2r)=r-4 equation: 



(r^2+5r-6)/4r+(3/2r)=r-4

We move all terms to the left:

(r^2+5r-6)/4r+(3/2r)-(r-4)=0



Domain of the equation: 4r!=0

r!=0/4

r!=0

r∈R





Domain of the equation: 2r)!=0

r!=0/1

r!=0

r∈R



We add all the numbers together, and all the variables

(r^2+5r-6)/4r+(+3/2r)-(r-4)=0

We get rid of parentheses

(r^2+5r-6)/4r+3/2r-r+4=0

We calculate fractions

We do not support erpression: r^3

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