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

May 25, 2018

$x = \frac{7}{18} \pm \frac{\sqrt{157}}{18}$

#### Explanation:

Given:

$y = 9 {x}^{2} - 7 x - 3$

Note that this is in standard form:

$y = a {x}^{2} + b x + c$

with $a = 9$, $b = - 7$ and $c = - 3$

It has zeros which we can find using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- \left(\textcolor{b l u e}{- 7}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 7}\right)}^{2} - 4 \left(\textcolor{b l u e}{9}\right) \left(\textcolor{b l u e}{- 3}\right)}}{2 \left(\textcolor{b l u e}{9}\right)}$

$\textcolor{w h i t e}{x} = \frac{7 \pm \sqrt{49 + 108}}{18}$

$\textcolor{w h i t e}{x} = \frac{7 \pm \sqrt{157}}{18}$

$\textcolor{w h i t e}{x} = \frac{7}{18} \pm \frac{\sqrt{157}}{18}$

Note that $157$ is prime, so $\sqrt{157}$ is already in simplest form.
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