Solve 0(x-3)^2+(y+2)^2=9

$$0 ( x - 3 ) ^ { 2 } + ( y + 2 ) ^ { 2 } = 9$$

Solve for y

$$y=-5$$

$$y=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0\left(x^{2}-6x+9\right)+\left(y+2\right)^{2}=9$$

Anything times zero gives zero.

$$0+\left(y+2\right)^{2}=9$$

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

$$0+y^{2}+4y+4=9$$

Add 0 and 4 to get 4.

$$4+y^{2}+4y=9$$

Subtract 9 from both sides.

$$4+y^{2}+4y-9=0$$

Subtract 9 from 4 to get -5.

$$-5+y^{2}+4y=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y^{2}+4y-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-4±\sqrt{4^{2}-4\left(-5\right)}}{2}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$y=\frac{-4±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$y=\frac{-4±\sqrt{36}}{2}$$

Take the square root of 36.

$$y=\frac{-4±6}{2}$$

Now solve the equation y=\frac{-4±6}{2} when ± is plus. Add -4 to 6.

$$y=\frac{2}{2}$$

Divide 2 by 2.

$$y=1$$

Now solve the equation y=\frac{-4±6}{2} when ± is minus. Subtract 6 from -4.

$$y=-\frac{10}{2}$$

Divide -10 by 2.

$$y=-5$$

The equation is now solved.

$$y=1 y=-5$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$0\left(x^{2}-6x+9\right)+\left(y+2\right)^{2}=9$$

Anything times zero gives zero.

$$0+\left(y+2\right)^{2}=9$$

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

$$0+y^{2}+4y+4=9$$

Add 0 and 4 to get 4.

$$4+y^{2}+4y=9$$

Subtract 4 from both sides.

$$y^{2}+4y=9-4$$

Subtract 4 from 9 to get 5.

$$y^{2}+4y=5$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+4y+2^{2}=5+2^{2}$$

Square 2.

$$y^{2}+4y+4=5+4$$

Add 5 to 4.

$$y^{2}+4y+4=9$$

Factor y^{2}+4y+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$y+2=3 y+2=-3$$

Subtract 2 from both sides of the equation.

$$y=1 y=-5$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

$$y=-5\text{ or }y=1$$

Solve for x

$$x\in \mathrm{R}$$

$$y=1\text{ or }y=-5$$