Question 977261
{{{x^2-2ax+a^2-(x^2-2bx+b^2)=0}}}
{{{-2ax+a^2+2bx-b^2=0}}}
{{{2bx-2ax+a^2-b^2=0}}}
{{{(2b-2a)x+a^2-b^2=0}}}
{{{(2b-2a)x=b^2-a^2}}}
{{{x=(b^2-a^2)/(2b-2a)}}}


a and b may be expected constants and unequal to eachother...
{{{x=(b-a)(b+a)/(2(b-a))}}}
and so this step should be justifiable:
{{{highlight(x=(b+a)/2)}}}