Question 1164770
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<pre>

In a coordinate plane, the points on ​the  x-axis  are the points (x,0) for arbitrary "x".


So, the points you are looking for, satisfy this distance equation


    {{{sqrt((x-2)^2 + (0-(-3))^2)}}} = 6,

or

    {{{sqrt((x-2)^2 + 9)}}} = 36.


Square both sides to get


    (x-2)^2 + 9 = 6^2

    (x-2)^2 = 36 - 9

    (x-2)^2 = 27

    x - 2 = {{{sqrt(27)}}} = +/- {{{3*sqrt(3)}}}


Hence, there are two values for x  {{{x[1]}}} = {{{2+3*sqrt(3)}}}  and  {{{x[2]}}} = {{{2-3*sqrt(3)}}}.


So, the points are  A = ({{{2+3*sqrt(3)}}},0)  and  B = ({{{2-3*sqrt(3)}}},0).     <U>ANSWER</U>
</pre>

Solved.