How do I solve this with the quadratic equation?

#2x^2-8x-24#

2 Answers
Feb 23, 2018

#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#.

Feb 23, 2018

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#.