Answer
$$6,8,10$$
Work Step by Step
Recall the Pythagorean Theorem, $a^2+b^2=c^2$
where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse.
If the sides of the triangle are three consecutive even integers, the triangle can be represented by: $a=x$, $b=(x+2)$, and $c=(x+4)$
$x^2 + (x+2)^2 = (x+4)^2$
$x^2 + x^2 +4x+4 = x^2+8x+16$
Combine like terms all on one side in standard quadratic equation form:
$ax^2 + bx +c=0$
$x^2-4x-12=0$
now, apply the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
where $a=1$, $b=-4$, and $c=-12$
$x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(-12)}}{2(1)}$
$x=\dfrac{4\pm\sqrt{16+48}}{2}$
$x=\dfrac{4\pm\sqrt{64}}{2}$
$x=\dfrac{4\pm8}{2}$
$x=\dfrac{4+8}{2}$ or $x=\dfrac{4-8}{2}$
$x=\dfrac{12}{2}$ or $x=\dfrac{-4}{2}$
$x=6$ or $x=-2$
Since the length of a side can not be negative, we know that $x=6$
Therefore the sides of the triangle are:
$6, (6+2), (6+4)$
$$6,8,10$$
