Question 812693
What if you show the equation as {{{sqrt((x-5)^2)=7-2x}}} ?

Square both sides:
{{{(x-5)^2=49-28x+4x^2}}}
{{{x^2-10x+25=4x^2-28x+49}}}
{{{-10x+25=3x^2-28x+49}}}
{{{25=3x^2-18x+49}}}
{{{3x^2-18x+49-25=0}}}
{{{3x^2-18x+24=0}}} then divide all by 3.
{{{x^2-6x+8=0}}}


*Use general solution to quadratic formula; the quadratic appears nonfactorable.
discrimt is 36-4*8=4
square root of 4 is 2.


{{{x=(6-2)/2}}} or {{{x=(6+2)/2}}}
{{{x=2}}} or {{{x=4}}}



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*The quadratic actually IS FACTORABLE.
{{{(x-2)(x-4)}}}
{{{x^2-2x-4x+8}}}
{{{x^2-6x+8}}}