Answer
$(4+3\sqrt{3},0) $ and $ (4-3\sqrt{3},0)$
Work Step by Step
A point on the x axis has coordinates $P_{1}(x,0).$
Given $P_{2}(4,-3)$, we find $x$ from
$\begin{aligned}
& \\
d(P_{1},P_{2}) & =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\
6 & =\sqrt{(4-x)^{2}+(-3-0)^{2}}\\
6^{2} & =(4-x)^{2}+(-3-0)^{2}\\
36 & =16-8x+x^{2}+9\\
0 & =x^{2}-8x-11
\end{aligned}$
Using the quadratic formula,
$x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=$
$=\displaystyle \frac{8\pm\sqrt{64+44}}{2}=$
$=\displaystyle \frac{8\pm\sqrt{108}}{2}$
$=\displaystyle \frac{8\pm 6\sqrt{3}}{2}$
$=4\pm 3\sqrt{3}$
The points have coordinates $(4+3\sqrt{3},0) $ and $ (4-3\sqrt{3},0)$
