Solve (x-3)(y+3)-(y-4)^2=-5

$$( x - 3 ) ( y + 3 ) - ( y - 4 ) ^ { 2 } = - 5$$

Solve for x

$$x=-\frac{-y^{2}+5y-20}{y+3}$$

$$y\neq -3$$

Steps for Solving Linear Equation

Use the distributive property to multiply x-3 by y+3.

$$xy+3x-3y-9-\left(y-4\right)^{2}=-5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$xy+3x-3y-9-\left(y^{2}-8y+16\right)=-5$$

To find the opposite of y^{2}-8y+16, find the opposite of each term.

$$xy+3x-3y-9-y^{2}+8y-16=-5$$

Combine -3y and 8y to get 5y.

$$xy+3x+5y-9-y^{2}-16=-5$$

Subtract 16 from -9 to get -25.

$$xy+3x+5y-25-y^{2}=-5$$

Subtract 5y from both sides.

$$xy+3x-25-y^{2}=-5-5y$$

Add 25 to both sides.

$$xy+3x-y^{2}=-5-5y+25$$

Add -5 and 25 to get 20.

$$xy+3x-y^{2}=20-5y$$

Add y^{2} to both sides.

$$xy+3x=20-5y+y^{2}$$

Combine all terms containing x.

$$\left(y+3\right)x=20-5y+y^{2}$$

The equation is in standard form.

$$\left(y+3\right)x=y^{2}-5y+20$$

Divide both sides by y+3.

$$\frac{\left(y+3\right)x}{y+3}=\frac{y^{2}-5y+20}{y+3}$$

Dividing by y+3 undoes the multiplication by y+3.

$$x=\frac{y^{2}-5y+20}{y+3}$$

Solve for y (complex solution)

$$y=\frac{-\sqrt{x^{2}+22x-55}+x+5}{2}$$

$$y=\frac{\sqrt{x^{2}+22x-55}+x+5}{2}$$

Solve for y

$$y=\frac{-\sqrt{x^{2}+22x-55}+x+5}{2}$$

$$y=\frac{\sqrt{x^{2}+22x-55}+x+5}{2}\text{, }x\geq 4\sqrt{11}-11\text{ or }x\leq -4\sqrt{11}-11$$