Solve (x+1/2)^2+(y-2/2)=9/4

$$( x + 1 / 2 ) ^ { 2 } + ( y - \frac { 2 } { 2 } ) = \frac { 9 } { 4 }$$

Solve for y

$$y=3-x-x^{2}$$

Solution Steps

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

$$x^{2}+x+\frac{1}{4}+y-\frac{2}{2}=\frac{9}{4}$$

Divide 2 by 2 to get 1.

$$x^{2}+x+\frac{1}{4}+y-1=\frac{9}{4}$$

Subtract 1 from \frac{1}{4} to get -\frac{3}{4}.

$$x^{2}+x-\frac{3}{4}+y=\frac{9}{4}$$

Subtract x^{2} from both sides.

$$x-\frac{3}{4}+y=\frac{9}{4}-x^{2}$$

Subtract x from both sides.

$$-\frac{3}{4}+y=\frac{9}{4}-x^{2}-x$$

Add \frac{3}{4} to both sides.

$$y=\frac{9}{4}-x^{2}-x+\frac{3}{4}$$

Add \frac{9}{4} and \frac{3}{4} to get 3.

$$y=3-x^{2}-x$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{13-4y}-1}{2}$$

$$x=\frac{\sqrt{13-4y}-1}{2}$$

Solve for x

$$x=\frac{-\sqrt{13-4y}-1}{2}$$

$$x=\frac{\sqrt{13-4y}-1}{2}\text{, }y\leq \frac{13}{4}$$