/ / / ____\\ / / ____\\\ / / / ____\\ / / ____\\\ / / / ____\\ / / ____\\\ / / / ____\\ / / ____\\\
| ___ 4 ___ |atan\\/ 47 /| ___ 4 ___ |atan\\/ 47 /|| | ___ 4 ___ |atan\\/ 47 /| ___ 4 ___ |atan\\/ 47 /|| | ___ 4 ___ |atan\\/ 47 /| ___ 4 ___ |atan\\/ 47 /|| | ___ 4 ___ |atan\\/ 47 /| ___ 4 ___ |atan\\/ 47 /||
|x + \/ 2 *\/ 3 *cos|------------| + I*\/ 2 *\/ 3 *sin|------------||*|x + \/ 2 *\/ 3 *cos|------------| - I*\/ 2 *\/ 3 *sin|------------||*|x + - \/ 2 *\/ 3 *cos|------------| + I*\/ 2 *\/ 3 *sin|------------||*|x + - \/ 2 *\/ 3 *cos|------------| - I*\/ 2 *\/ 3 *sin|------------||
\ \ 2 / \ 2 // \ \ 2 / \ 2 // \ \ 2 / \ 2 // \ \ 2 / \ 2 //
$$\left(x + \left(\sqrt{2} \sqrt[4]{3} \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)} - \sqrt{2} \sqrt[4]{3} i \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)}\right)\right) \left(x + \left(\sqrt{2} \sqrt[4]{3} \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)} + \sqrt{2} \sqrt[4]{3} i \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)}\right)\right) \left(x + \left(- \sqrt{2} \sqrt[4]{3} \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)} + \sqrt{2} \sqrt[4]{3} i \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)}\right)\right) \left(x + \left(- \sqrt{2} \sqrt[4]{3} \cos{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)} - \sqrt{2} \sqrt[4]{3} i \sin{\left(\frac{\operatorname{atan}{\left(\sqrt{47} \right)}}{2} \right)}\right)\right)$$
(((x + sqrt(2)*3^(1/4)*cos(atan(sqrt(47))/2) + i*sqrt(2)*3^(1/4)*sin(atan(sqrt(47))/2))*(x + sqrt(2)*3^(1/4)*cos(atan(sqrt(47))/2) - i*sqrt(2)*3^(1/4)*sin(atan(sqrt(47))/2)))*(x - sqrt(2)*3^(1/4)*cos(atan(sqrt(47))/2) + i*sqrt(2)*3^(1/4)*sin(atan(sqrt(47))/2)))*(x - sqrt(2)*3^(1/4)*cos(atan(sqrt(47))/2) - i*sqrt(2)*3^(1/4)*sin(atan(sqrt(47))/2))
The perfect square
Let's highlight the perfect square of the square three-member
$$\left(y^{4} - y^{2}\right) + 12$$
To do this, let's use the formula
$$a y^{4} + b y^{2} + c = a \left(m + y^{2}\right)^{2} + n$$
where
$$m = \frac{b}{2 a}$$
$$n = \frac{4 a c - b^{2}}{4 a}$$
In this case
$$a = 1$$
$$b = -1$$
$$c = 12$$
Then
$$m = - \frac{1}{2}$$
$$n = \frac{47}{4}$$
So,
$$\left(y^{2} - \frac{1}{2}\right)^{2} + \frac{47}{4}$$