Find the point on the X axis whose distance from the point (-7,5) is 13 units.

Dear Student,
Let the point be P(x,0) and the other point be Q(-7,5), PQ=13 units  Then according to distance formula,
$$\begin{gathered} PQ=\sqrt{\left(x_1-x_2\right){ }^2+ {\left(y_1-y_2\right)}^2} \\ 13=\sqrt{{\left(x+7\right)}^2+(0-5{)}^2} \\ Squaring both the sides \\ {\left(13\right)}^2={\left(x+7\right)}^2+(0-5{)}^2 \\ 169=x^2+14x+49+25 \\ 169=x^2+14x+74 \\ {x}^2+14x-95=0 \\ Splitting the middle term \\ {x}^2+19x-5x-95=0 \\ x(x+19)-5(x+19)=0 \\ (x-5\left)\right(x+19)=0 \\ x=5 OR x=-19 \end{gathered}$$
The point is (5,0) or (-19,0)
Hope this information has solved your queries 
Regards