Adding a microlitre of HCl into water; what is pH?

use the hendersen hasselbach equation

Mind demonstrating?

pH of H2O = 7, therefore:

pH = -log[H+]
7 = -log[H+]
1 x 10-7 = [H+]
Since this is one liter of water, then you will have 1 x 10-7 mol [H+]

If we add 100 x 10-6 (100 microL) of 0.002M HCl, then we are going to add:
1 x 10-4 L x 2 x 10-3 mol/L HCl = 2 x 10-7 mol H+  (since HCl will completely dissociate).

Sum the two molar amounts of [H+], and you get 3.00 x 10-7 mol H+. 
However, to get the concentration of H+, we must know the volume:
1 L + 100 x 10-6 = 1.0001 L of solution.

Now then, we know the moles of H+ and the volume of the solution, so we can calculate the molarity:
3.00 x 10-7 mol H+ / 1.0001 L of solution = 3.00 x 10-7 M H+

Thusly,
pH = -log[H+] = -log[3.00 x 10-7 H+] = 6.52

Let me know if this is right.

0.002 M × 0.0001 L = 2 × 10-7 mol
The answer is 6.62 and apparently they used quadratic formula with after solving for expression:

H2O = H+ + OH-
            x         x
           2*10^-7

(x)(x+(2*10^-7)) = 10^-14

But my question is how though? When I try to solve this quadratic I am getting 0....

Who is "they"?

I got 6.52 just like Doctor-2-B, using the pH equation method above.

Also, that random quadratic that "they" gave obviously has two solutions that aren't zero.

As an aside, the topic says you're adding a microliter of water, but in fact, you're adding 100 \({\mu}L\).  That may be kinda confusing?