Question 1020925
The quickest way of solving this equation is to square both sides.  This will effectively take away the absolute value bars.

{{{(x-4)^2 = (2x+1)^2}}}
<==> {{{x^2 - 8x +16 = 4x^2 + 4x +1}}}
<==> {{{0 = 3x^2 +12x -15}}}

<==> {{{0 = x^2 +4x -5}}} ==> (x+5)(x-1) = 0 ==>x = {{{highlight(-5)}}} or {{{highlight(1)}}}.
Direct substitution of each value into the original equation verifies that both of them are solutions.

There is another way of solving this equation which requires the determination of the critical numbers of the equation (effectively partitioning the real number line) to eliminate the absolute value bars.