Solve y_{12}=2x^2+2b-2

$$y _ { 12 } = 2 x ^ { 2 } + 2 b - 2$$

Solve for b

$$b=\frac{2+y_{12}-2x^{2}}{2}$$

Steps for Solving Linear Equation

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

$$2x^{2}+2b-2=y_{12}$$

Subtract 2x^{2} from both sides.

$$2b-2=y_{12}-2x^{2}$$

Add 2 to both sides.

$$2b=y_{12}-2x^{2}+2$$

The equation is in standard form.

$$2b=2+y_{12}-2x^{2}$$

Divide both sides by 2.

$$\frac{2b}{2}=\frac{2+y_{12}-2x^{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$b=\frac{2+y_{12}-2x^{2}}{2}$$

Divide y_{12}-2x^{2}+2 by 2.

$$b=-x^{2}+\frac{y_{12}}{2}+1$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{2\left(y_{12}-2b+2\right)}}{2}$$

$$x=\frac{\sqrt{2\left(y_{12}-2b+2\right)}}{2}$$

Solve for x

$$x=\frac{\sqrt{2\left(y_{12}-2b+2\right)}}{2}$$

$$x=-\frac{\sqrt{2\left(y_{12}-2b+2\right)}}{2}\text{, }y_{12}\geq 2b-2$$