Solve -(t+1/4)^2=9/16

$$- ( t + \frac { 1 } { 4 } ) ^ { 2 } = \frac { 9 } { 16 }$$

Solve for t

$$t=-\frac{1}{4}-\frac{3}{4}i=-0.25-0.75i$$

$$t=-\frac{1}{4}+\frac{3}{4}i=-0.25+0.75i$$

Steps by Finding Square Root

Dividing by -1 undoes the multiplication by -1.

$$\left(t+\frac{1}{4}\right)^{2}=\frac{\frac{9}{16}}{-1}$$

Divide \frac{9}{16} by -1.

$$\left(t+\frac{1}{4}\right)^{2}=-\frac{9}{16}$$

Take the square root of both sides of the equation.

$$t+\frac{1}{4}=\frac{3}{4}i t+\frac{1}{4}=-\frac{3}{4}i$$

Subtract \frac{1}{4} from both sides of the equation.

$$t+\frac{1}{4}-\frac{1}{4}=\frac{3}{4}i-\frac{1}{4} t+\frac{1}{4}-\frac{1}{4}=-\frac{3}{4}i-\frac{1}{4}$$

Subtracting \frac{1}{4} from itself leaves 0.

$$t=\frac{3}{4}i-\frac{1}{4} t=-\frac{3}{4}i-\frac{1}{4}$$

Subtract \frac{1}{4} from \frac{3}{4}i.

$$t=-\frac{1}{4}+\frac{3}{4}i$$

Subtract \frac{1}{4} from -\frac{3}{4}i.

$$t=-\frac{1}{4}-\frac{3}{4}i$$

The equation is now solved.

$$t=-\frac{1}{4}+\frac{3}{4}i t=-\frac{1}{4}-\frac{3}{4}i$$