How do you find the zeros of $y = x^2 - 5x -13$ using the quadratic formula?

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Answer 1 · 2018-03-13T17:18:13

The quadratic formula always gives two answers. In this case, $x=6.8875$ and $x=-1.8875$

To find the 'zeros', you want to solve the function for when y=0:

$0=x^2-5x-13$

There is a general form for this, which is what we will be using to describe the quadratic formula:

$0=ax^2+bx+c$

The quadratic formula looks like this:

$(-b+-sqrt(b^2-4ac))/(2a)$

For this case:
$a=1$
$b=-5$
$c=-13$

Now, let's plug those into the formula:

$x=(-(-5)+-sqrt((-5)^2-4(1)(-13)))/(2(1))$

$x=(5+-sqrt(25-4*(-13)))/2$

$x=(5+-sqrt(25+52))/2 rArr x=(5+-sqrt(77))/2$

Now, we solve for both scenarios of x:

$color(red)(x=(5+sqrt(77))/2 rArr x=6.8875$

$color(blue)(x=(5-sqrt(77))/2 rArr x=-1.8875$

How do you find the zeros of y = x^2 - 5x -13 y = x^2 - 5x -13 using the quadratic formula?