<s>[INST] Find complex roots of quartic function $(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$ I found a math problem involving complex number

Find all complex number z such that
  $$(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2$$

The complex number form is z = a + bi
If I multiply all the factor to make it to the standard form of a polynomial, It would be too long to solve. Not even substituting z = a + bi to the polynomial yet.
I don't think that's a good way to solve
I want to know if there's a simpler way to solve this problem, can anyone shows me a hint? 
 [/INST] Let $z=\frac{x}{12}.$ 
Thus, $$(3z + 1)(4z + 1)(6z + 1)(12z + 1)-2=$$
$$=\left(\frac{x}{4}+1\right)\left(\frac{x}{3}+1\right)\left(\frac{x}{2}+1\right)(x+1)-2=$$
$$=\frac{1}{24}((x+4)(x+3)(x+2)(x+1)-48)=$$
$$=\frac{1}{24}((x^2+5x+4)(x^2+5x+6)-48)=$$
$$=\frac{1}{24}(x^2+5x+12)(x^2+5x-2)=$$
$$=(12z^2+5z+1)(72z^2+30z-1).$$
Can you end it now? 
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