inequalities

You need to set 2 cases for the range of value of X in order to solve the problem.

For x>6
(x-6)^2 = 3(x-6) + 10 (since x-6 is positive when x>6)
x^2 - 12x + 36 - 3x + 18 - 10 = 0
x^2 + 24x + 44 = 0
x = -2(rejected)or x= -22 (rejected)

For x<6
(x-6)^2 = -3(x-6) + 10 (since x-6 is negative when x<6)
x^2 - 12x + 36 + 3x - 18 - 10 = 0
x^2 -9x + 8 = 0
x=8 (rejected) or x=1

So,the root of the equation "(x-6)^2 = 3|x-6|+10" is x=1

Similarly
For x>6
(x-6)^2 - 3(x-6) - 10 < 0
x^2 + 24x + 44 < 0
x < -2 (rejected)or x > -22
by combing x>-22 and x>6
the first range of x is x > -22

For x<6
(x-6)^2 + 3(x-6) - 10 < 0
x^2 - 12x + 36 + 3x - 18 - 10 < 0
x^2 -9x + 8 < 0
x < 8 or x > 1
The second range of x is x < 6
Range of values of x： -22 < x < 6

Here is a simpler method:

For (x - 6)^2 = 3|x - 6|+10, note that (x - 6)^2 = |x - 6|^2
So, we have
|x - 6|^2 = 3|x - 6|+10
|x - 6|^2 - 3|x - 6| - 10 = 0
(Note that y^2 - 3y - 10 = 0 gives y = 5 or -2)
|x - 6| = 5 or - 2 (rejected, as absolute value always non-negative)
x - 6 = 5 or -5
x = 11 or 1

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for (x-6)^2 < 3|x-6|+10, we have
|x - 6|^2 - 3|x - 6| - 10 < 0
(Note that y^2 - 3y - 10 < 0 gives -2 < y < 5)
-2 < |x - 6| < 5
|x - 6| < 5 (again, sbsolute value always non-negative)
-5 < x - 6 < 5
-1 < x < 11