Suppose a, b denote the distinct real roots of the quadratic polynomial $$x^{2} + 20x − 2020$$ and
suppose c, d denote the distinct complex roots of the quadratic polynomial $$x^{2} − 20x + 2020$$. Then the value of
ac(a - c) + ad(a - d) + bc(b - c) + bd(b + d)
is

Let $$a,\;b$$ be the (real) roots of $$x^{2}+20x-2020=0$$ and $$c,\;d$$ be the (complex-conjugate) roots of $$x^{2}-20x+2020=0$$.

For the first quadratic, by Vieta’s relations
$$a+b=-20,\qquad ab=-2020.$$

For the second quadratic,
$$c+d=20,\qquad cd=2020.$$

We have to evaluate
$$S=ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d).$$

Expand each term:
$$\begin{aligned} ac(a-c)&=a^{2}c-ac^{2},\\ ad(a-d)&=a^{2}d-ad^{2},\\ bc(b-c)&=b^{2}c-bc^{2},\\ bd(b-d)&=b^{2}d-bd^{2}. \end{aligned}$$

Add the four expressions:
$$\begin{aligned} S&=(a^{2}+b^{2})(c+d)-(a+b)(c^{2}+d^{2}).\qquad -(1) \end{aligned}$$

First compute $$a^{2}+b^{2}$$:
$$(a^{2}+b^{2})=(a+b)^{2}-2ab=(-20)^{2}-2(-2020)=400+4040=4440.$$

Next compute $$c^{2}+d^{2}$$:
$$c^{2}+d^{2}=(c+d)^{2}-2cd=20^{2}-2(2020)=400-4040=-3640.$$

Substitute these values in $$(1)$$:
$$\begin{aligned} S&=(4440)(20)-(-20)(-3640)\\[4pt] &=88800-72800\\[4pt] &=16000. \end{aligned}$$

Hence the required value is $$16000$$.

Option E which is: 16000