Solve y=(t+1/4)^2+2

$$y = ( \frac { t + 1 } { 4 } ) ^ { 2 } + 2$$

Solve for y

$$y=\frac{\left(t+1\right)^{2}+32}{16}$$

Short Solution Steps

To raise \frac{t+1}{4} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{4^{2}}{4^{2}}.

$$y=\frac{\left(t+1\right)^{2}}{4^{2}}+\frac{2\times 4^{2}}{4^{2}}$$

Since \frac{\left(t+1\right)^{2}}{4^{2}} and \frac{2\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$y=\frac{\left(t+1\right)^{2}+2\times 4^{2}}{4^{2}}$$

Do the multiplications in \left(t+1\right)^{2}+2\times 4^{2}.

$$y=\frac{t^{2}+2t+1+32}{4^{2}}$$

Combine like terms in t^{2}+2t+1+32.

$$y=\frac{t^{2}+2t+33}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$y=\frac{t^{2}+2t+33}{16}$$

Divide each term of t^{2}+2t+33 by 16 to get \frac{1}{16}t^{2}+\frac{1}{8}t+\frac{33}{16}.

$$y=\frac{1}{16}t^{2}+\frac{1}{8}t+\frac{33}{16}$$

Solve for t (complex solution)

$$t=-4\sqrt{y-2}-1$$

$$t=4\sqrt{y-2}-1$$

Solve for t

$$t=-4\sqrt{y-2}-1$$

$$t=4\sqrt{y-2}-1\text{, }y\geq 2$$