Solve left(t-4right)^2=left(t+4right)^2+32

$${ \left(t-4 \right) }^{ 2 } = { \left(t+4 \right) }^{ 2 } +32$$

Solve for t

$$t=-2$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-4\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+4\right)^{2}.

$$t^{2}-8t+16=t^{2}+8t+16+32$$

Add 16 and 32 to get 48.

$$t^{2}-8t+16=t^{2}+8t+48$$

Subtract t^{2} from both sides.

$$t^{2}-8t+16-t^{2}=8t+48$$

Combine t^{2} and -t^{2} to get 0.

$$-8t+16=8t+48$$

Subtract 8t from both sides.

$$-8t+16-8t=48$$

Combine -8t and -8t to get -16t.

$$-16t+16=48$$

Subtract 16 from both sides.

$$-16t=48-16$$

Subtract 16 from 48 to get 32.

$$-16t=32$$

Divide both sides by -16.

$$t=\frac{32}{-16}$$

Divide 32 by -16 to get -2.

$$t=-2$$