Question 707008
At the point where 2 circles are (mutually) tangent their tangents are the same line, perpendicular to the radius of each circle.
The radii of the 2 circles form a continuous segment from one circle center to the other, and is one of the sides of the triangle in the problem.
So, the lengths of the sides of your triangle are:
{{{14+15=29}}}
{{{14+x}}} and
{{{15+x}}}
 
It looks like
{{{drawing(300,300,-20,45,-20,45,
triangle(0,0,20,0,0,21),circle(0,21,15),
circle(0,0,6),circle(20,0,14)
)}}} if the legs (shorter sides) of the right triangle measure {{{14+x}}} and {{{15+x}}}
If the longest side is the one measuring {{{15+x}}},
then I will not draw the whole largest circle {{{drawing(450,80,-20,430,-20,60,
triangle(0,0,420,0,0,29),circle(0,29,15),
circle(0,0,14),circle(420,0,406)
)}}}
 
In the first case,
{{{(15+x)^2+(14+x)^2=29^2}}} --> {{{225+30x+x^2+196+28x+x^2=841}}} --> {{{2x^2+58x+421=841}}} --> {{{2x^2+58x-420=0}}} --> {{{x^2+29x+210=0}}}
If you are good at factoring, you factor the quadratic into
{{{(x-6)(x+35)=0}}} and find the positive solution you need as {{{highlight(x=6)}}}
If you are not that good at factoring, the quadratic formula will give you the same answer.
 
In the second case,
{{{(15+x)^2=(14+x)^2+29^2}}} --> {{{225+30x+x^2=196+28x+x^2+841}}} --> {{{225+30x=11037+28x}}} --> {{{2x=812}}} --> {{{highlight(x=406)}}}
 
It could have a few more answers, but I have to leave for my paying job now.
The problem most likely assumes that each circle is outside the other two.
(Circles can also be internally tangent, with one circle inside the other and just touching at one point. That would complicate the problem and give you extra solutions, but few people would think of internally tangent circles when reading or writing a problem).