x^2+x-6/8x+4-3x^2+6x-72/8x+4=x+3/4x+2

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Solution for x^2+x-6/8x+4-3x^2+6x-72/8x+4=x+3/4x+2 equation: 



x^2+x-6/8x+4-3x^2+6x-72/8x+4=x+3/4x+2

We move all terms to the left:

x^2+x-6/8x+4-3x^2+6x-72/8x+4-(x+3/4x+2)=0



Domain of the equation: 8x!=0

x!=0/8

x!=0

x∈R





Domain of the equation: 4x+2)!=0

x∈R



We add all the numbers together, and all the variables

-2x^2+7x-6/8x-72/8x-(x+3/4x+2)+8=0

We get rid of parentheses

-2x^2+7x-6/8x-72/8x-x-3/4x-2+8=0

We calculate fractions


-2x^2+7x-x+(-288x-6)/32x^2+(-24x)/32x^2-2+8=0

We add all the numbers together, and all the variables

-2x^2+6x+(-288x-6)/32x^2+(-24x)/32x^2+6=0

We multiply all the terms by the denominator


-2x^2*32x^2+6x*32x^2+(-288x-6)+(-24x)+6*32x^2=0

Wy multiply elements

-64x^4+192x^3+192x^2+(-288x-6)+(-24x)=0

We get rid of parentheses

-64x^4+192x^3+192x^2-288x-24x-6=0

We do not support expression: x^4

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