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

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Answer 1 · 2016-06-14T09:05:04

The zeros of the given Quadratic Polynomial are $(43+-sqrt(2725))/6.$

Formula to find the zeros, say $alpha$ & $beta,$ of a Quadratic Polynomial $y=P(x)=ax^2+bx+c$ is

$alpha,beta=(-b+-sqrt(b^2-4ac))/(2a)$

In our case, $a=3, b=-43,c=-73.$
$:. alpha,beta= {43+-sqrt(1849+876)}/6=(43+-sqrt(2725))/6.$

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

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