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

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

Solve for k

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

$$x\neq -1$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by k-3.

$$kx^{2}+\left(2k-6\right)x+k-3=0$$

Use the distributive property to multiply 2k-6 by x.

$$kx^{2}+2kx-6x+k-3=0$$

Add 6x to both sides. Anything plus zero gives itself.

$$kx^{2}+2kx+k-3=6x$$

Add 3 to both sides.

$$kx^{2}+2kx+k=6x+3$$

Combine all terms containing k.

$$\left(x^{2}+2x+1\right)k=6x+3$$

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

$$\frac{\left(x^{2}+2x+1\right)k}{x^{2}+2x+1}=\frac{6x+3}{x^{2}+2x+1}$$

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

$$k=\frac{6x+3}{x^{2}+2x+1}$$

Divide 6x+3 by x^{2}+2x+1.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{-k+\sqrt{3\left(3-k\right)}+3}{k}\text{; }x=\frac{-k-\sqrt{3\left(3-k\right)}+3}{k}\text{, }&k\neq 0\\x=-\frac{1}{2}\text{, }&k=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-k+\sqrt{3\left(3-k\right)}+3}{k}\text{; }x=\frac{-k-\sqrt{3\left(3-k\right)}+3}{k}\text{, }&k\neq 0\text{ and }k\leq 3\\x=-\frac{1}{2}\text{, }&k=0\end{matrix}\right.$$