Recall the definition of the *absolute value*:

#|C|=C# for all #C>=0# and

#|C|=-C# for all #C<0#.

Let's divide the real numbers into two separate areas:

A1 is a set of all real numbers #x#, for which #x^2-2x-16 >= 0#.

A2 is a set of all real numbers #x#, for which #x^2-2x-16 < 0#.

Let's determine these areas. Graphically, #x^2-2x-16# is represented by a parabola with its two "horns" directed upwards. Therefore, it is greater than #0# to the left of the left (smaller) solution of an equation #x^2-2x-16=0# and to the right of the right (bigger) solution of this equation. In between the solutions this quadratic polynomial is less than #0#.

graph{x^2-2x-16 [-10, 10, -25, 25]}

The two solutions to the equation #x^2-2x-16=0# are

#x_(1,2)=(2+-sqrt(4+64))/2=1+-sqrt(17)#

So, area A1 where our quadratic polynomial is non-negative consists of two non-intersecting parts:

#x <= 1-sqrt(17)# and

#x >= 1+sqrt(17)#.

Area A2 where our quadratic polynomial is negative is characterized by a combined inequality

#1-sqrt(17) < x < 1+sqrt(17)#

**Case 1**:

#x <= 1-sqrt(17)# OR #x >= 1+sqrt(17)#

Since #sqrt(17)~=4.123#, our area, approximately, consists of two parts:

#x <= -3.123# OR #x >= 5.123#

Then our quadratic polynomial is non-negative and we can simply drop the absolute value sign obtaining an equation

#x^2-2x-16=8# or, equivalently, #x^2-2x-24=0#.

Its solutions are

#x_(1,2)=(2+-sqrt(4+96))/2=(2+-10)/2#, that is

#x_1=6#, #x_2=-4#

Solution #x_1=6# is valid since #6 >= 5.123#.

Solution #x_2=-4# is valid since #-4 <= -3.123#.

**Case 2**:

#1-sqrt(17) < x < 1+sqrt(17)#

Approximately, it means

#-3.123 < x < 5.123#

Then our quadratic polynomial is negative and, if we want to get rid of absolute value sign, we have to change the sign of this polynomial getting the equation

#-(x^2-2x-16)=8# or, equivalently, #x^2-2x-8=0#.

Its solutions are

#x_(3,4)=(2+-sqrt(4+32))/2=(2+-6)/2#, that is

#x_3=4#, #x_4=-2#

Solution #x_3=4# is valid since #-3.123 < 4 < 5.123#.

Solution #x_4=-2# is valid since #-3.123 < -2 < 5.123#.

**Solution**:

#x_1 = 6#

#x_2 = -4#

#x_3= 4#

#x_4 = -2#

Graphically, the absolute value of a given polynomial looks like

graph{|x^2-2x-16| [-10, 10, -25, 25]}

If you draw a line #y=8# on this graph, it will intersect the graph at four points listed above as solutions.