How do I solve this with the quadratic equation?
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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#.
I have #x = 6#, and #x = -2#
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#.
