Solve x^2+(2-2a)x+1=0

$$x ^ { 2 } + ( 2 - 2 a ) x + 1 = 0$$

Solve for a

$$a=\frac{\left(x+1\right)^{2}}{2x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2-2a by x.

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

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

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

Subtract 2x from both sides.

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

Subtract 1 from both sides.

$$-2ax=-x^{2}-2x-1$$

The equation is in standard form.

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

Divide both sides by -2x.

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

Dividing by -2x undoes the multiplication by -2x.

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

Divide -\left(x+1\right)^{2} by -2x.

$$a=\frac{\left(x+1\right)^{2}}{2x}$$

Solve for x (complex solution)

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

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

Solve for x

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

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