Answer
The side lengths of the triangle are 6, 8 and 10.
Work Step by Step
$$x^2+(x+2)^2=(x+4)^2$$
$$x^2+x^2+4x+4=x^2+8x+16$$
$$2x^2+4x+4=x^2+8x+16$$
$$x^2-4x-12=0$$
$$(x-6)(x+2)=0$$
$$x=6,-2$$
Length can not be negative, so x=6.
The side lengths of the triangle are 6, 8 and 10.