Solve (k-5)x^2-4kx+(k-2)=0

$$( k - 5 ) x ^ { 2 } - 4 k x + ( k - 2 ) = 0$$

Solve for k

$$k=\frac{5x^{2}+2}{x^{2}-4x+1}$$

$$x\neq \sqrt{3}+2\text{ and }x\neq 2-\sqrt{3}$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-5 by x^{2}.

$$kx^{2}-5x^{2}-4kx+k-2=0$$

Add 5x^{2} to both sides. Anything plus zero gives itself.

$$kx^{2}-4kx+k-2=5x^{2}$$

Add 2 to both sides.

$$kx^{2}-4kx+k=5x^{2}+2$$

Combine all terms containing k.

$$\left(x^{2}-4x+1\right)k=5x^{2}+2$$

Divide both sides by x^{2}-4x+1.

$$\frac{\left(x^{2}-4x+1\right)k}{x^{2}-4x+1}=\frac{5x^{2}+2}{x^{2}-4x+1}$$

Dividing by x^{2}-4x+1 undoes the multiplication by x^{2}-4x+1.

$$k=\frac{5x^{2}+2}{x^{2}-4x+1}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(k-1\right)\left(3k+10\right)}+2k}{k-5}\text{; }x=\frac{-\sqrt{\left(k-1\right)\left(3k+10\right)}+2k}{k-5}\text{, }&k\neq 5\\x=\frac{3}{20}\text{, }&k=5\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(k-1\right)\left(3k+10\right)}+2k}{k-5}\text{; }x=\frac{-\sqrt{\left(k-1\right)\left(3k+10\right)}+2k}{k-5}\text{, }&\left(k\neq 5\text{ and }k\geq 1\right)\text{ or }k\leq -\frac{10}{3}\\x=\frac{3}{20}\text{, }&k=5\end{matrix}\right.$$