If the perimeter is 24, then the sum of the other two sides is 14.

We know that both legs are less than 10, because the hypotenuse is the longest side.

So, let's let A be one leg, and let's let the other leg be 14-A.

We know the Pythagorean theorem must hold, so

A^{2} + (14-A)^{2} = 100

FOIL

(14-A)^{2 }= 196 - 28A + A^{2}

Then combine like terms

2A^{2} -28A + 196 = 100

Looks like everything's even, so we can divide by 2 to help factor later. And let's get the constants on the same side while we're at it.

A^{2} - 14A + 48 = 0

We need two numbers that multiply to 48 and add to -14, that's -6 and -8.

So we factor down to (A-6)(A-8) = 0

So it looks like we have a 6,8,10 triangle

Let me know if you need this explained differently, or if you just want a part of it restated.