How do you solve #( x - 6)^2 = 19#?

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Answer 1 · 2016-04-14T04:49:23

#x=6+sqrt19, 6-sqrt 19#

#(x-6)^2=19#

Take the square root of both sides.

#x-6=+-sqrt 19#

Add #6# to both sides of the equation.

#x=6+-sqrt 19#

Solve for #x#.

#x=6+sqrt19, 6-sqrt 19#