# How do you find the zeros, real and imaginary, of y= 3x^2-43x-73 using the quadratic formula?

Jun 14, 2016

The zeros of the given Quadratic Polynomial are $\frac{43 \pm \sqrt{2725}}{6.}$

#### Explanation:

Formula to find the zeros, say $\alpha$ & $\beta ,$ of a Quadratic Polynomial $y = P \left(x\right) = a {x}^{2} + b x + c$ is

$\alpha , \beta = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our case, $a = 3 , b = - 43 , c = - 73.$
$\therefore \alpha , \beta = \frac{43 \pm \sqrt{1849 + 876}}{6} = \frac{43 \pm \sqrt{2725}}{6.}$

We note that since $\Delta = {b}^{2} - 4 a c = 2725 > 0$, the given quadratic can not have any complex zero.