Question 1014683
{{{x^2-1=8x}}}
using method of completing the square
{{{x^2-8x= 1}}}
add 16 to both sides
{{{x^2-8x+16}}} = {{{1 +16}}}
{{{x^2-8x+16}}} = {{{17}}}
{{{(x-4)^2}}} = {{{17}}}
{{{x-4}}} = +-{{{ sqrt(17)}}}
{{{x}}} = {{{- sqrt(17)+4}}} or {{{+ sqrt(17)+4}}}
{{{x}}} = {{{-4.123+4}}} or {{{4.123+4}}}
{{{x}}} = {{{-0.123}}} or {{{8.123}}}