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

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Solution for -4(2x^2+3)-4x=7x-8(x^2+6x-3^2) equation:



-4(2x^2+3)-4x=7x-8(x^2+6x-3^2)
We move all terms to the left:
-4(2x^2+3)-4x-(7x-8(x^2+6x-3^2))=0
We add all the numbers together, and all the variables
-4x-4(2x^2+3)-(7x-8(x^2+6x-3^2))=0
We multiply parentheses
-8x^2-4x-(7x-8(x^2+6x-3^2))-12=0
We calculate terms in parentheses: -(7x-8(x^2+6x-3^2)), so:
7x-8(x^2+6x-3^2)
We multiply parentheses
-8x^2+7x-48x+72
We add all the numbers together, and all the variables
-8x^2-41x+72
Back to the equation:
-(-8x^2-41x+72)
We get rid of parentheses
-8x^2+8x^2+41x-4x-72-12=0
We add all the numbers together, and all the variables
37x-84=0
We move all terms containing x to the left, all other terms to the right
37x=84
x=84/37
x=2+10/37

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