Solve t^2=(t-1)^2+4

$$t ^ { 2 } = ( t - 1 ) ^ { 2 } + 4$$

Solve for t

$$t = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-1\right)^{2}.

$$t^{2}=t^{2}-2t+1+4$$

Add 1 and 4 to get 5.

$$t^{2}=t^{2}-2t+5$$

Subtract t^{2} from both sides.

$$t^{2}-t^{2}=-2t+5$$

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

$$0=-2t+5$$

Swap sides so that all variable terms are on the left hand side.

$$-2t+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-2t=-5$$

Divide both sides by -2.

$$t=\frac{-5}{-2}$$

Fraction \frac{-5}{-2} can be simplified to \frac{5}{2} by removing the negative sign from both the numerator and the denominator.

$$t=\frac{5}{2}$$