Would you use Ferrari-Cardano formula?

$\begin{aligned} x^4 - 6x^3 + 11x^2 - 6x - 2020 & = 0 \\ x^4 - 6x^3 + 11x^2 - 6x + 1 & = 2021 \\ (x^2 - 3x+1)^2 & = 2021 \\ \implies x^2 - 3x + 1 & = \pm \sqrt{2021} \end{aligned}$

$\implies \begin{cases} x^2 - 3x + 1 - \sqrt{2021} = 0 & \implies x = \frac {3 \pm \sqrt{9-4 + 4\sqrt{2021}}}2 = \frac {3 \pm \sqrt{5+4\sqrt{2021}}}2 = x_1, x_2 & \small \blue{\text{real roots}} \\ x^2 - 3x + 1 + \sqrt{2021} = 0 & \implies x = \frac {3 \pm \sqrt{9-4 - 4\sqrt{2021}}}2 = \frac {3 \pm i\sqrt{4\sqrt{2021}-5}}2 = x_3, x_4 & \small \red{\text{complex roots}} \end{cases}$

$\implies \left(\left(\frac {x_1-x_2}2\right)^2 + \left(\frac {|x_3-x_4|}2\right)^2\right)^2 = \left(\frac {5+4\sqrt{2021}}4 + \frac {4\sqrt{2021}-5}4 \right)^2 = \boxed{8084}$

We are going to make $c=2020$ . Then the polynomial $f(x)$ can be represented as $f(x)=x^4-6x^3+11x^2-6x-c$
Now we can make $x=X+\frac{3}{2}.$ Then $f(X+\frac{3}{2})=(X+\frac{3}{2})^4-6(X+\frac{3}{2})^3+11(X+\frac{3}{2})^2-6(X+\frac{3}{2})-c =X^4-\frac{5}{2}X^2+\frac{9}{16}-c.$ Let us solve the equation $X^4-\frac{5}{2}X^2+\frac{9}{16}-c=0.$ The solutions of this biquadratic equations are: $\sqrt{\sqrt{1+c}-\frac{5}{4}}, -\sqrt{\sqrt{1+c}-\frac{5}{4}}, i \sqrt{\sqrt{1+c}+\frac{5}{4}}, -i \sqrt{\sqrt{1+c}+\frac{5}{4}}.$ Then the roots of the polynomial $f(x)$ are $x_1=\frac{3}{2}+\sqrt{\sqrt{1+c}-\frac{5}{4}}, x_2=\frac{3}{2}-\sqrt{\sqrt{1+c}-\frac{5}{4}}, x_3= \frac{3}{2}+i \sqrt{\sqrt{1+c}+\frac{5}{4}}, x_4=\frac{3}{2}-i \sqrt{\sqrt{1+c}+\frac{5}{4}}.$ Therefore, $((\frac{x_1-x_2}{2})^2 +(\frac{|x_3-x_4|}{2})^2)^2=(\sqrt{1+c}-\frac{5}{4}+\sqrt{1+c}+\frac{5}{4})^2=(2\sqrt{1+c})^2=4(1+c)=\boxed{8084}.$

We might recognise the coefficients of the quartic - it can be rewritten as $f(x)=x(x-1)(x-2)(x-3)-M$

where $M=2020$ . If we replace $x$ with $x+k$ for some constant $k$ , the quantity we're asked to find will not change (it's invariant under translations). Noting that $x(x-1)(x-2)(x-3)$ is symmetric about $x=\frac32$ suggests putting $k=\frac32$ ; so let $g(x)=f\left(x+\frac32\right)=\left(x+\frac32\right)\left(x+\frac12\right)\left(x-\frac12\right)\left(x-\frac32\right)-M$

This becomes $g(x)=\left(x^2-\frac94\right)\left(x^2-\frac14\right)-M=x^4-\frac52 x^2+\frac{9}{16}-M$

We'll factor this as two quadratics; one with roots $x_1,x_2$ and one with roots $x_3,x_4$ . Note that $g(x)$ contains no odd powers of $x$ ; the factorisation will be of the form $\frac{1}{16}\left(x^2-p\right)\left(x^2-q\right)$ , where $p+q=\frac52$ and $pq=\frac{9}{16}-M$ , and without loss of generality $p>0>q$ .

The roots are, of course, $x_{1,2}=\pm \sqrt{p}$ and $x_{3,4}=\pm \sqrt{q}$ ; so $\begin{aligned} \left(\left(\frac{x_1-x_2}{2}\right)^2+\left(\frac{\left|x_3-x_4\right|}{2}\right)^2\right)^2 &= \left(p-q\right)^2 \\ &=(p+q)^2-4pq \\ &=\frac{25}{4}-\frac{9}{4}+4M \\ &=\boxed{8084} \end{aligned}$