Question 59439
(x=2)^2=16  You mis-typed this.  I'm going to make the first = a -.  If you meant +, you can probably figure out how to solve it from this any way.
{{{(x-2)^2=16}}}
{{{sqrt((x-2)^2)}}}=+\-{{{sqrt(16)}}}
x-2=+\-4
x-2=2+\-4
x=2-4 and x=2+4
x=-2 and x=6
Happy Calculating!!!