Question

How do I solve this with the quadratic equation?

`2x^2-8x-24`

Answer 1

`x=6` and `x=-2`

Explanation:

This equation is in standard form, `ax^2+bx+c`. The Quadratic Formula states that the roots to this quadratic are at:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

In our equation, `a=2, b= -8` and `c=-24`. Now, let's plug in:

`x=(8+-sqrt((-8)^2-4(2)(-24)))/(2(2))`

The terms in the radical simplify to:

`x=(8+-sqrt(64-8(-24)))/4`

`x=(8+-sqrt(64+192))/4`

`x=(8+-sqrt(256))/4`

`x=(8+-16)/4`

We can factor out a `4` (because all of the terms are divisible by `4`). We get:

`x=(2+-4)/1`

Now, let's break up our equations:

`x=2+4` `=>x=6`

`x=2-4` `=>x=-2`

Our zeroes are `x=6` and `x=-2`.

Answer 2

I have `x = 6`, and `x = -2`

Explanation:

The quadratic equation is as follows:

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

All we have to do is plug the values in and solve.

`a=2, b=-8, c=-24`

So, when we plug the numbers in, we get:

`(-(-8)+-sqrt((-8)^2-4(2)(-24)))/(2(2)`

Now we can start simplifying.

`(8+-sqrt(64+192))/4`

`(8+-sqrt(256))/4`

`(8+-16)/4`

So our answers will be `x = 6` and `x = -2`.