Question 30996
{{{4/(x-2)=1+6/(x+2)}}} replace the 1 with {{{(x+2)/(x+2)}}}
{{{4/(x-2)=(x+2)/(x+2)+6/(x+2)}}} combine the comon fractions
{{{4/(x-2)=(x+2+6)/(x+2)}}} multiply both sides by (x-2) to cancel out the x-2 on the left, also 2+6=8
{{{4=((x+8)(x-2))/(x+2)}}} now multiply both sides by x+2 to cancel out the x+2 on the right
{{{4*(x+2)=(x+8)(x-2)}}} distribute the left side
{{{4x+8=x^2+6x-16}}} subtract both the 4x and the 8 from both sides
{{{0=x^2+2x-32}}}
Now using the quadratic formula:  {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
where {{{Ax^2+Bx+C=0}}}
we  would get
{{{x=4.745}}} and {{{x=-6.745}}}