Solution:
Given, p(x) = ax^2 + bx + c and ac = b
Let the roots of the quadratic equation be α and β.

Sum of the roots, α + β = -b/a
Product of the roots, αβ = c/a

To find one of the zeroes, let's assume that α = k
Then, β = c/(ak)

As ac = b, we get c = b/a

Substituting the values of α and β, we get
k + c/(ak) = -b/a
ak^2 + c = -bk/a

Multiplying both sides by a, we get
ak^2 + ac = -bk

Substituting c = b/a, we get
ak^2 + b = -bk

Simplifying further, we get
ak^2 + bk + b = 0

This is a quadratic equation in k, which can be solved using the quadratic formula.

The discriminant of the equation is b^2 - 4ab = b(b - 4a)

As ac = b, we have a = b/c
Substituting this in the discriminant, we get
b(b - 4b/c) = b^2(1 - 4/c)

Since a, b, and c are all non-zero, we can say that c ≠ 0
Therefore, 1 - 4/c ≠ 0

Hence, the discriminant is non-zero, which means that the quadratic equation has two distinct real roots.

Therefore, one of the zeroes is k, which is given by the quadratic formula as
k = (-b + sqrt(b^2 - 4ab))/2a or k = (-b - sqrt(b^2 - 4ab))/2a

Simplifying this, we get
k = (-b ± sqrt(b^2 - 4ac))/2a

Therefore, one of the zeroes is given by
α = k = (-b + sqrt(b^2 - 4ac))/2a or α = k = (-b - sqrt(b^2 - 4ac))/2a

As we are only interested in one of the zeroes, we can take the negative root, which gives us
α = k = (-b - sqrt(b^2 - 4ac))/2a

Substituting the value of c = b/a, we get
α = k = (-b - sqrt(b^2 - 4ab/a))/2a

Simplifying this, we get
α = k = (-b - sqrt((b^2a - 4ab)/a))/2a
α = k = (-b - sqrt(b^2 - 4a))/2a

Therefore, one of the zeroes is given by
α = (-b - sqrt(b^2 - 4a))/2a, which is option (C).

Hence, the correct answer is option (C).