# How do you find the zeros, real and imaginary, of y= 23x^2+18x-24 using the quadratic formula?

Mar 27, 2018

There are two real roots:

$- \frac{9}{23} - \frac{\sqrt{633}}{23}$ and $- \frac{9}{23} + \frac{\sqrt{633}}{23}$

#### Explanation:

We have:

$y = 23 {x}^{2} + 18 x - 24$

So the roots are given by the solution of the equation:

$23 {x}^{2} + 18 x - 24 = 0$

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

With $a = 23 , b = 18$ and $c = - 24$ we have:

$x = \frac{- 18 \pm \sqrt{{18}^{2} - 4 \left(23\right) \left(- 24\right)}}{2 \left(23\right)}$
$\setminus \setminus = \frac{- 18 \pm \sqrt{324 + 2208}}{46}$
$\setminus \setminus = \frac{- 18}{46} \pm \frac{\sqrt{2532}}{46}$
$\setminus \setminus = - \frac{9}{23} \pm \frac{\sqrt{633}}{23}$

Hence, there are two real roots:

$- \frac{9}{23} - \frac{\sqrt{633}}{23}$ and $- \frac{9}{23} + \frac{\sqrt{633}}{23}$