How do you solve #x^2 + 6x = 1#?

Question

6751 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T06:14:55

#x=-3+-sqrt10#
#x approx 0.16228 or approx-6.16228#

#x^2+6x=1#

#x^2+6x-1=0#

#x= (-6+-sqrt(6^2-4xx1xx(-1)))/(2xx1)#

#= -3+-sqrt40/2#

#= -3 +-sqrt(2xx2xx2xx5)/2#

#=-3+- (2sqrt10)/2#

#=-3+-sqrt10#

#x=-3+-sqrt10#

#x approx 0.16228 or approx-6.16228#