Solve (t-4)^2=(t+4)^2+3=

$$( t - 4 ) ^ { 2 } = ( t + 4 ) ^ { 2 } + 3 =$$

Solve for t

$$t=-\frac{3}{16}=-0.1875$$

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}.

$$t^{2}-8t+16=\left(t+4\right)^{2}+3$$

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

$$t^{2}-8t+16=t^{2}+8t+16+3$$

Add 16 and 3 to get 19.

$$t^{2}-8t+16=t^{2}+8t+19$$

Subtract t^{2} from both sides.

$$t^{2}-8t+16-t^{2}=8t+19$$

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

$$-8t+16=8t+19$$

Subtract 8t from both sides.

$$-8t+16-8t=19$$

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

$$-16t+16=19$$

Subtract 16 from both sides.

$$-16t=19-16$$

Subtract 16 from 19 to get 3.

$$-16t=3$$

Divide both sides by -16.

$$t=\frac{3}{-16}$$

Fraction \frac{3}{-16} can be rewritten as -\frac{3}{16} by extracting the negative sign.

$$t=-\frac{3}{16}$$