Solve 2x^2+2y+4=1

$$2 x ^ { 2 } + 2 y + 4 = 1$$

Solve for y

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

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides.

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

Subtract 4 from both sides.

$$2y=1-2x^{2}-4$$

Subtract 4 from 1 to get -3.

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

The equation is in standard form.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -3-2x^{2} by 2.

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{-4y-6}}{2}$$

$$x=\frac{\sqrt{-4y-6}}{2}$$

Solve for x

$$x=\frac{\sqrt{-4y-6}}{2}$$

$$x=-\frac{\sqrt{-4y-6}}{2}\text{, }y\leq -\frac{3}{2}$$