Given the function h(x) = 3x2 and the condition h(a+1) = 3, find the possible values of a.

We are given the function $h(x) = 3x^2$.
To find $h(a+1)$, we replace $x$ with $(a+1)$:

$$h(a+1) = 3(a+1)^2$$

We are told that $h(a+1) = 3$. So, we can set the expression from Step 1 equal to 3:

$$3(a+1)^2 = 3$$

Divide both sides by 3:

$$(a+1)^2 = \frac{3}{3}$$ $$(a+1)^2 = 1$$

Take the square root of both sides. Remember to consider both positive and negative roots:

$$a+1 = \pm\sqrt{1}$$ $$a+1 = \pm 1$$

This gives two possible cases:

Case 1: $a+1 = 1$
Subtract 1 from both sides:

$$a = 1 - 1$$ $$a = 0$$

Case 2: $a+1 = -1$
Subtract 1 from both sides:

$$a = -1 - 1$$ $$a = -2$$

The values of $a$ are 0 and $-2$.

The value of $a$ is $\boxed{0 \text{ or } -2}$.