Once you know the value of $h(8)$, set $h(t)$ equal to that same value and write an equation involving $t$.

After you set up the equation in $t$, get all terms on one side, factor, and solve for both possible $t$ values.

You will get two solutions for $t$. Use the condition $t \neq 8$ to decide which solution is allowed.

In Desmos, type y = x^2 - 8x + 12 to graph the function $h(x)$.

Either add a table for the function and include $x=8$, or type a separate line x^2 - 8x + 12 with x = 8 substituted (e.g., 8^2 - 8*8 + 12) to see the output value of $h(8)$.

On a new line, enter the equation x^2 - 8x + 12 = (the value you found for $h(8)$). Desmos will show the solutions for $x$ that make this true; identify both $x$-values and then choose the one that is not $8$ to match the condition in the problem.

We are given

To use the condition $h(8)=h(t)$, first find $h(8)$:

So $h(8)$ equals $12$.

Now use the condition $h(8)=h(t)$. Since $h(8)=12$, we need $h(t)$ to also equal $12$:

But $h(t) = t^2 - 8t + 12$, so we have

Solve the equation

Subtract $12$ from both sides:

Now factor the left side:

So the equation becomes

From

either $t = 0$ or $t - 8 = 0$, which means $t = 8$.

The problem states $t \neq 8$, so we must choose the other solution.

Therefore, the value of $t$ is $0$.