3(x+2)(x-1))/2-(2x+3)(x+4)/7=0

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Solution for 3(x+2)(x-1))/2-(2x+3)(x+4)/7=0 equation: 



3(x+2)(x-1))/2-(2x+3)(x+4)/7=0



Domain of the equation: 2-(2x!=0

We move all terms containing x to the left, all other terms to the right

-(2x!=-2

x∈R



We multiply parentheses ..

3(+x^2-1x+2x-2))/2-(2x+3)(x+4)/7=0

We calculate fractions


3)(x+(21x^2+21x)/(-(2x*7+2)+4*2-2x/(-(2x*7+2)=0



We calculate terms in parentheses: +(21x^2+21x)/(-(2x*7+2)+4*2-2x/(-(2x*7+2), so:

21x^2+21x)/(-(2x*7+2)+4*2-2x/(-(2x*7+2

determiningTheFunctionDomain
21x^2+21x)/(-(2x*7+2)-2x/(-(2x*7+2+4*2

We get rid of parentheses

21x^2+21x)/(-2x*7-2x/(-(2x*7+2+4*2-2

We calculate fractions


21x^2-2x*7+21x*(-(2x*7+2+4*2-2)/((*(-(2x*7+2+4*2-2)+(-2x*()/((*(-(2x*7+2+4*2-2)

We can not solve this equation

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