Question 972335
{{{abs(x^2+6) = 5x}}}...........by definition of abs value,{{{abs(x^2+6)=x^2+6}}}

{{{x^2+6 = 5x}}}

{{{x^2 -5x+6=0}}}

{{{x^2-2x -3x+6=0}}}

{{{(x^2-2x)-(3x-6)=0}}}

{{{x(x-2)-3(x-2)=0}}}

{{{(x-3)(x-2)=0}}}

solutions:

if {{{(x-3)=0}}}=>{{{x=3}}}

if {{{(x-2)=0}}}=>{{{x=2}}}