Does Anyone Know the Answer to This Difficult Question in Calculus

Hello everyone,
I just came up with a question while I was playing with microsoft excel and I came up with an interesting question:

Assume a the following function,
f(x) = x^2-a*x,
where "a" is an arbitrary constant. The function passes the x-axis at a certain point from x= 0 to 20.

It is known in calculus that the area between the curve and the x-axis could be found by integrating the function and evaluating the integral from the start to the end points. If we wanted to assign a certain value for "a" such that the area before of the graph up to the intersection of f(x) with the x-axis, is one half the area after the intersection, i.e., the area between the curve and the x-axis from x=0 to the point where the function hit the graph, is one half the area from that same point to 20.

What should the value of "a" be?

I honestly don't know the answer, but from graphing the function at different values of "a" I found that it is solvable. I just need to know how to solve it analytically.

Thank you everyone for your time!

f(x) = x^2 - ax
f(x) = x(x - a)
0 = x(x - a)

x = 0 or a

So x = 0 HAS to be one of the places it crosses the x axis. And we must also have that:

0 < a ≤ 20

In order for the graph to also cross the x axis at some point between 0 and 20. Integrate f(x) from 0 to a:

∫ [x^2 - ax] dx from 0 to a
(1/3)x^3 - (a/2)x^2 from 0 to a
(1/3)a^3 - (a/2)a^2 <-- note that the lower limit of zero will be zero,
(1/3)a^3 - (1/2)a^3
(-1/6)a^3

Now evaluate from a to 20 instead:

(1/3)x^3 - (a/2)x^2 from a to 20
(1/3)(20)^3 - (a/2)(20)^2 - [(1/3)(a)^3 - (a/2)(a)^2]
(1/6)a^3 - 200a + 8000/3

Now we want the first area to be half the second:

(-1/6)a^3 = [1/2][(1/6)a^3 - 200a + 8000/3]

We might need some absolute value brackets here because the part before the intersection is going to be negative: