1+(x2-5x-24)/3x=(x-6)/3x

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Solution for 1+(x2-5x-24)/3x=(x-6)/3x equation: 



1+(x2-5x-24)/3x=(x-6)/3x

We move all terms to the left:

1+(x2-5x-24)/3x-((x-6)/3x)=0



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R





Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(+x^2-5x-24)/3x-((x-6)/3x)+1=0

We calculate fractions


(x^2-5x-24)*3x)/9x^2+(-((x-6)*3x)/9x^2+1=0

We calculate fractions


((x^2-5x-24)*3x)*9x^2)/(9x^2+(*9x^2)+(-((x-6)*3x)*9x^2)/(9x^2+(*9x^2)+1=0

We get rid of parentheses

((x^2-5x-24)*3x)*9x^2)/(9x^2+*9x^2+(-((x-6)*3x)*9x^2)/(9x^2+(*9x^2)+1=0

We calculate fractions


*9x^2+(((x^2-5x-24)*3x)*9x^2)*(9x^2+*9x^2+1)/((9x^2*(9x^2+(*9x^2)+1)+((-((x-6)*3x)*9x^2)*9x^2/((9x^2*(9x^2+(*9x^2)+1)=0



We calculate terms in parentheses: +(((x^2-5x-24)*3x)*9x^2)*(9x^2+*9x^2+1)/((9x^2*(9x^2+(*9x^2)+1)+((-((x-6)*3x)*9x^2)*9x^2/((9x^2*(9x^2+(*9x^2)+1), so:

((x^2-5x-24)*3x)*9x^2)*(9x^2+*9x^2+1)/((9x^2*(9x^2+(*9x^2)+1)+((-((x-6)*3x)*9x^2)*9x^2/((9x^2*(9x^2+(*9x^2)+1

We can not solve this equation

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