Solve 2x^2+2y^2+6x-4y-k=0

$$2 x ^ { 2 } + 2 y ^ { 2 } + 6 x - 4 y - k = 0$$

Solve for k

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

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides. Anything subtracted from zero gives its negation.

$$2y^{2}+6x-4y-k=-2x^{2}$$

Subtract 2y^{2} from both sides.

$$6x-4y-k=-2x^{2}-2y^{2}$$

Subtract 6x from both sides.

$$-4y-k=-2x^{2}-2y^{2}-6x$$

Add 4y to both sides.

$$-k=-2x^{2}-2y^{2}-6x+4y$$

The equation is in standard form.

$$-k=4y-2y^{2}-6x-2x^{2}$$

Divide both sides by -1.

$$\frac{-k}{-1}=\frac{4y-2y^{2}-6x-2x^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k=\frac{4y-2y^{2}-6x-2x^{2}}{-1}$$

Divide -2x^{2}-2y^{2}-6x+4y by -1.

$$k=2x^{2}+6x+2y^{2}-4y$$

Solve for x (complex solution)

$$x=\frac{\sqrt{9+2k+8y-4y^{2}}-3}{2}$$

$$x=\frac{-\sqrt{9+2k+8y-4y^{2}}-3}{2}$$

Solve for x

$$x=\frac{\sqrt{9+2k+8y-4y^{2}}-3}{2}$$

$$x=\frac{-\sqrt{9+2k+8y-4y^{2}}-3}{2}\text{, }k\geq 2y^{2}-4y-\frac{9}{2}$$