Question 330528
{{{x^2 + 52 = 8x}}}
{{{x^2-8x + 52 = 0}}}
{{{x^2-8x+16+36=0}}}
{{{(x-4)^2=-36}}}
{{{x-4=sqrt(-36)}}}
{{{x-4=0 +- 6i}}} where {{{i=sqrt(-1)}}}
{{{ x=4 +- 6i}}}