## College Algebra (6th Edition)

$x\ne-16,12$
To determine the numbers that must be excluded from the domain, we determine the values that make the denominator equal to zero. Solve the absolute value by removing the absolute value signs and solving, and also solve it by removing the absolute value signs and solving when the other side is negative. $|x+2|-14=0$ $|x+2|=14$ $x+2=14$ $x=12$ $x+2=-14$ $x=-16$ $x\ne-16,12$