Solve (t+1/2)^2-2(t+1/2)t+t^2

$$( t + \frac { 1 } { 2 } ) ^ { 2 } - 2 ( t + \frac { 1 } { 2 } ) t + t ^ { 2 }$$

Evaluate

$$\frac{1}{4}=0.25$$

Solution Steps

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

$$t^{2}+t+\frac{1}{4}-2\left(t+\frac{1}{2}\right)t+t^{2}$$

Use the distributive property to multiply -2 by t+\frac{1}{2}.

$$t^{2}+t+\frac{1}{4}+\left(-2t-1\right)t+t^{2}$$

Use the distributive property to multiply -2t-1 by t.

$$t^{2}+t+\frac{1}{4}-2t^{2}-t+t^{2}$$

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

$$-t^{2}+t+\frac{1}{4}-t+t^{2}$$

Combine t and -t to get 0.

$$-t^{2}+\frac{1}{4}+t^{2}$$

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

$$\frac{1}{4}$$

Factor

$$\frac{1}{2 ^ {2}} = 0.25$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{\left(2t+1\right)^{2}-4\left(2t+1\right)t+4t^{2}}{4}$$