Solve (t-1.1)^2=(t-2.1)^2

$$( t - 1.1 ) ^ { 2 } = ( t - 2.1 ) ^ { 2 }$$

Solve for t

$$t=1.6$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-1.1\right)^{2}.

$$t^{2}-2.2t+1.21=\left(t-2.1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-2.1\right)^{2}.

$$t^{2}-2.2t+1.21=t^{2}-4.2t+4.41$$

Subtract t^{2} from both sides.

$$t^{2}-2.2t+1.21-t^{2}=-4.2t+4.41$$

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

$$-2.2t+1.21=-4.2t+4.41$$

Add 4.2t to both sides.

$$-2.2t+1.21+4.2t=4.41$$

Combine -2.2t and 4.2t to get 2t.

$$2t+1.21=4.41$$

Subtract 1.21 from both sides.

$$2t=4.41-1.21$$

Subtract 1.21 from 4.41 to get 3.2.

$$2t=3.2$$

Divide both sides by 2.

$$t=\frac{3.2}{2}$$

Expand \frac{3.2}{2} by multiplying both numerator and the denominator by 10.

$$t=\frac{32}{20}$$

Reduce the fraction \frac{32}{20} to lowest terms by extracting and canceling out 4.

$$t=\frac{8}{5}$$