How do you solve #x^ { 2} - 8x = - 17#?

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Answer 1 · 2017-06-21T09:20:06

#x= 4+-i#

#x^2-8x =-17#

#x^2-8x+17=0#

#x= (8+-sqrt(8^2-4xx1xx17))/(2xx1)#

#= (8+-sqrt(64-68))/2#

#= (8+- sqrt(-4))/2#

#= (8+-sqrt4 * i)/2#

#= (8+-2i)/2#

#= 4+-i#