2(x-4)^2+7(x-4)+5=0

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Solution for 2(x-4)^2+7(x-4)+5=0 equation: 



2(x-4)^2+7(x-4)+5=0

We multiply parentheses

2(x-4)^2+7x-28+5=0

We add all the numbers together, and all the variables

7x+2(x-4)^2-23=0

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

7x+2(x-4)^2=23

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