(x-3)3x2+4x+3=(x-3)x2+x-5

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Solution for (x-3)3x2+4x+3=(x-3)x2+x-5 equation: 



(x-3)3x^2+4x+3=(x-3)x2+x-5

We move all terms to the left:

(x-3)3x^2+4x+3-((x-3)x2+x-5)=0

We add all the numbers together, and all the variables

4x+(x-3)3x^2-((x-3)x2+x-5)+3=0



We calculate terms in parentheses: -((x-3)x2+x-5), so:

(x-3)x2+x-5

We add all the numbers together, and all the variables

x+(x-3)x2-5

We multiply parentheses

x^3-3x^2+x-5

We do not support expression: x^3

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