# What are all the real zeros of y=(x-12)^3-10?

Aug 20, 2017

The real zero is:

$x = 12 + \sqrt[3]{10}$

#### Explanation:

Given:

$y = {\left(x - 12\right)}^{3} - 10$

The real zero is:

$x = 12 + \sqrt[3]{10}$

The other zeros are both non-real complex:

$x = 12 + \omega \sqrt[3]{10}$

$x = 12 + {\omega}^{2} \sqrt[3]{10}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive complex cube root of $1$.