Solve (x-2)^2+(y+3)=4

$$( x - 2 ) ^ { 2 } + ( y + 3 ) = 4$$

Solve for y

$$y=-\left(x-3\right)\left(x-1\right)$$

Solution Steps

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

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

Add 4 and 3 to get 7.

$$x^{2}-4x+7+y=4$$

Subtract x^{2} from both sides.

$$-4x+7+y=4-x^{2}$$

Add 4x to both sides.

$$7+y=4-x^{2}+4x$$

Subtract 7 from both sides.

$$y=4-x^{2}+4x-7$$

Subtract 7 from 4 to get -3.

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

Solve for x (complex solution)

$$x=-\sqrt{1-y}+2$$

$$x=\sqrt{1-y}+2$$

Solve for x

$$x=-\sqrt{1-y}+2$$

$$x=\sqrt{1-y}+2\text{, }y\leq 1$$