Solve (k-1)x^2-2kx+k-3=0

$$( k - 1 ) x ^ { 2 } - 2 k x + k - 3 = 0$$

Solve for k

$$k=\frac{x^{2}+3}{\left(x-1\right)^{2}}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-1 by x^{2}.

$$kx^{2}-x^{2}-2kx+k-3=0$$

Add x^{2} to both sides. Anything plus zero gives itself.

$$kx^{2}-2kx+k-3=x^{2}$$

Add 3 to both sides.

$$kx^{2}-2kx+k=x^{2}+3$$

Combine all terms containing k.

$$\left(x^{2}-2x+1\right)k=x^{2}+3$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)k}{x^{2}-2x+1}=\frac{x^{2}+3}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$k=\frac{x^{2}+3}{x^{2}-2x+1}$$

Divide x^{2}+3 by x^{2}-2x+1.

$$k=\frac{x^{2}+3}{\left(x-1\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{k+\sqrt{4k-3}}{k-1}\text{; }x=\frac{k-\sqrt{4k-3}}{k-1}\text{, }&k\neq 1\\x=-1\text{, }&k=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{k+\sqrt{4k-3}}{k-1}\text{; }x=\frac{k-\sqrt{4k-3}}{k-1}\text{, }&k\neq 1\text{ and }k\geq \frac{3}{4}\\x=-1\text{, }&k=1\end{matrix}\right.$$