Complex roots of a quadratic equation

The question:

Use the properties of the complex conjugate to show that if the complex number 'z' is a root of a quadratic equation $\displaystyle ax^2 + bx + c = 0$ with a, b, c being real coefficients, then so is the conjugate of 'z'.

My attempt:

Let z = $\displaystyle z_{1} \pm z_2i$

$\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = z_{1} \pm z_2i$

$\displaystyle \frac{1}{2a}(-b \pm \sqrt{b^2 - 4ac}) = z_{1} \pm z_2i$

Basically, I changed the quadratic formula to look like a complex number in Cartesian form. Is this an acceptable answer? My textbook doesn't have solutions for 'show' and 'prove' questions.

Thanks!