How do you solve #x^ { 2} - 8x - 78= 8#?

1 Answer
Shreya Naik
Jun 2, 2017

a little simplification needs to be done
ie, #x^2 - 8x-78=8#
#x^2-8x-86=0# ( bringing +8 to left side)

Explanation:

#x= -b +- sqrt(b^2-4ac)/2/a#
solve for x putting the values of a =+1 , b= -8 , c= - 86
and find the answer , it will give the most appropriate answer
........................hope it helps.......................:)

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