Solve x^2+2bx-2x+2-1=0

$$x ^ { 2 } + 2 b x - 2 x + 2 - 1 = 0$$

Solve for b

$$b=-\frac{\left(x-1\right)^{2}}{2x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 1 from 2 to get 1.

$$x^{2}+2bx-2x+1=0$$

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

$$2bx-2x+1=-x^{2}$$

Add 2x to both sides.

$$2bx+1=-x^{2}+2x$$

Subtract 1 from both sides.

$$2bx=-x^{2}+2x-1$$

The equation is in standard form.

$$2xb=-x^{2}+2x-1$$

Divide both sides by 2x.

$$\frac{2xb}{2x}=-\frac{\left(x-1\right)^{2}}{2x}$$

Dividing by 2x undoes the multiplication by 2x.

$$b=-\frac{\left(x-1\right)^{2}}{2x}$$

Solve for x (complex solution)

$$x=\sqrt{b\left(b-2\right)}-b+1$$

$$x=-\sqrt{b\left(b-2\right)}-b+1$$

Solve for x

$$x=\sqrt{b\left(b-2\right)}-b+1$$

$$x=-\sqrt{b\left(b-2\right)}-b+1\text{, }b\geq 2\text{ or }b\leq 0$$