#### Answer

$t=-2$

#### Work Step by Step

$(t-4)^{2}=(t+4)^2+32$
Let's remember the following properties:
$(a+b)^{2}=a^{2}+2ab+b^{2}$
$(a-b)^{2}=a^{2}-2ab+b^{2}$
Evaluate the powers present in the equations using said properties:
$t^{2}-2(t)(4)+4^{2}=t^{2}+2(t)(4)+4^{2}+32$
We can get rid of the terms present in both sides of the equation.
This means $t^{2}$ and $4^{2}$ will disappear from the equation.
We get:
$-2(t)(4)=2(t)(4)+32$
Simplify and solve for $t$:
$-8t=8t+32$
$-8t-8t=32$
$-16t=32$
$t=\dfrac{32}{-16}=-2$