# How do you find the zeros, real and imaginary, of  y=x^2/3+1/7x-5/9  using the quadratic formula?

Jan 25, 2016

$x = \frac{- 3 \pm \sqrt{\frac{1007}{3}}}{14}$

#### Explanation:

The quadratic formula states that the zeros of a quadratic function in the form $y = a {x}^{2} + b x + c$ are equal to

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

With the quadratic $y = {x}^{2} / 3 + \frac{1}{7} x - \frac{5}{9}$, we can see that

$\left\{\begin{matrix}a = \frac{1}{3} \\ b = \frac{1}{7} \\ c = - \frac{5}{9}\end{matrix}\right.$

Plug these into the quadratic formula:

$x = \frac{- \frac{1}{7} \pm \sqrt{{\left(\frac{1}{7}\right)}^{2} - \left(4 \times \frac{1}{3} \times - \frac{5}{9}\right)}}{2 \times \frac{1}{3}}$

Simplify.

$x = \frac{- \frac{1}{7} \pm \sqrt{\frac{1}{49} + \frac{20}{27}}}{\frac{2}{3}}$

$x = \frac{- \frac{1}{7} \pm \sqrt{\frac{27}{1323} + \frac{980}{1323}}}{\frac{2}{3}}$

$x = \frac{- \frac{1}{7} \pm \sqrt{\frac{1007}{1323}}}{\frac{2}{3}}$

$x = \frac{- \frac{1}{7} \pm \sqrt{\frac{1007}{9 \times 49 \times 3}}}{\frac{2}{3}}$

$x = \frac{- \frac{1}{7} \pm \frac{1}{21} \sqrt{\frac{1007}{3}}}{\frac{2}{3}}$

$x = \left(- \frac{1}{7} \pm \frac{1}{21} \sqrt{\frac{1007}{3}}\right) \cdot \left(\frac{3}{2}\right)$

$x = \frac{- \frac{3}{7} \pm \frac{1}{7} \sqrt{\frac{1007}{3}}}{2}$

$x = \frac{- 3 \pm \sqrt{\frac{1007}{3}}}{14}$