Question #34b64

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Answer 1 · 2017-05-15T01:35:38

#x=1#

First, add #7# on both sides so that it's in the form #ax^2+bx+c=0#

#7r^2-14r+7=0#

Now that we have it in the form, #ax^2+bx+c#, we can identify:

#a=7#
#b=-14#
#c=7#

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

Plug into the formula:

#x=(14+-sqrt(14^2-4(7)(7)))/(2(7))#

Solve inside the square root:

#x=(14+-sqrt(0))/(2(7))#

#sqrt(0)# is just #0#. So our answer is:

#x=14/14#

That is:

#x=1#