How do you solve #x^2 – 5x – 1 = - 7# using the quadratic formula?
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1 Answer
May 13, 2016
#x=(5+-sqrt(-23))/4#
It has imaginary roots
Explanation:
Given -
#x^2-5x-1=-7#
Take the constant to the LHS.
#2x^2-5x-1+7=0#
#2x^2-5x+6=0#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(-5) +- sqrt((-5)^2-(4xx2xx6)))/(2xx2)#
#x=(5+-sqrt(25-48))/4#
#x=(5+-sqrt(-23))/4#
It has imaginary roots
