Solve D=(t/2)^2-1

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

Solve for D

$$D=\frac{t^{2}-4}{4}$$

Short Solution Steps

To raise \frac{t}{2} to a power, raise both numerator and denominator to the power and then divide.

$$D=\frac{t^{2}}{2^{2}}-1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2^{2}}{2^{2}}.

$$D=\frac{t^{2}}{2^{2}}-\frac{2^{2}}{2^{2}}$$

Since \frac{t^{2}}{2^{2}} and \frac{2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$D=\frac{t^{2}-2^{2}}{2^{2}}$$

Do the multiplications in t^{2}-2^{2}.

$$D=\frac{t^{2}-4}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$D=\frac{t^{2}-4}{4}$$

Divide each term of t^{2}-4 by 4 to get \frac{1}{4}t^{2}-1.

$$D=\frac{1}{4}t^{2}-1$$

Solve for t

$$t=2\sqrt{D+1}$$

$$t=-2\sqrt{D+1}\text{, }D\geq -1$$