## Precalculus: Mathematics for Calculus, 7th Edition

$t=-2$
$(t-4)^{2}=(t+4)^2+32$ 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 $t$: $-8t=8t+32$ $-8t-8t=32$ $-16t=32$ $t=\dfrac{32}{-16}=-2$