Solve y=x^2/2+2c-2

$$y = \frac { x ^ { 2 } } { 2 } + 2 c - 2$$

Solve for c

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

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$\frac{x^{2}}{2}+2c-2=y$$

Subtract \frac{x^{2}}{2} from both sides.

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

Add 2 to both sides.

$$2c=y-\frac{x^{2}}{2}+2$$

Multiply both sides of the equation by 2.

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

The equation is in standard form.

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

Divide both sides by 4.

$$\frac{4c}{4}=\frac{4+2y-x^{2}}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$c=\frac{4+2y-x^{2}}{4}$$

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

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

Solve for x (complex solution)

$$x=-\sqrt{2\left(y-2c+2\right)}$$

$$x=\sqrt{2\left(y-2c+2\right)}$$

Solve for x

$$x=\sqrt{2\left(y-2c+2\right)}$$

$$x=-\sqrt{2\left(y-2c+2\right)}\text{, }y\geq 2c-2$$